TASK 2
Michal Kinel
2017-04-09
- 1 Carga de paquetes
- 2 Descarga y manejo de datos
- 3 Apartado 1
- 4 Apartado 2: Ajuste de los modelos CAPM para los activos
- 5 Apartado 3: los parámetros de los CAPM y los índices son constantes en el tiempo?
- 6 Apartado 4: problemas de heteroscedasticidad condicional?
- 7 Apartado 5: relación de cointegración?
1 Carga de paquetes
Comenzamos la tarea cargando los paquetes que serán necesarios.
library(quantmod)
library(PerformanceAnalytics)
library(dygraphs)
library(forecast)
library(FinTS)
library(rugarch)
library(gets)
library(ggplot2)
library(car)
library(astsa)
library(parallel)
library(tseries)
library(dynlm)
library(tseries)
library(forecast)
library(lmtest)
library(sandwich)
library(plotly)
library(tsoutliers)
library(urca)
library(gets)
library(dyn)
library(cointReg)
library(ggplot2)
library(zoo)
library(vars)2 Descarga y manejo de datos
Descargamos 3 valores del índice EUROSTOX50:
- ASML.AS- ASML Holding N.V.
- BAYN.DE- Bayer AG
- BNP.PA- BNP Paribas S.A.
getSymbols(c("^STOXX50E","ASML.AS", "BAYN.DE", "BNP.PA"), from= '2005-01-03' )## [1] "STOXX50E" "ASML.AS" "BAYN.DE" "BNP.PA"# descargar el tipo de interés
getSymbols("DTB3", src = "FRED", from= '2005-01-03' )## [1] "DTB3"Anualizamos los tipos de los Bonos Alemanes a 10 años:
DTB3 <- DTB3/(253)Seleccionamos los datos ajustasos y los unimos con merge en caso de ralizar tratamiento de datos conjunto:
fdata <- merge(STOXX50E$STOXX50E.Adjusted, ASML.AS$ASML.AS.Adjusted, BAYN.DE$BAYN.DE.Adjusted,BNP.PA$BNP.PA.Adjusted)
names(fdata) <- c("EUROSTOX", "ASML", "BAYN", "BNP")Por el otro lado, extraemos los datos por si fuera necesario tratamiento individual.
EUROSTOX <- fdata$EUROSTOX
ASML <- fdata$ASML
BAYN <- fdata$BAYN
BNP <- fdata$BNPVeamos la representación gráfica de los valores obtenidos:
dygraph(EUROSTOX, main = "EUROSTOX50") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(ASML, main = "ASML Holding N.V.") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(BAYN, main = "Bayer AG") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(BNP, main = "BNP Paribas S.A.") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))2.1 Cálculo de los retornos como porcentaje
rdata <- Return.calculate(fdata, method = "log")*100
rdata <- na.omit(merge(rdata, DTB3))
head(rdata)## EUROSTOX ASML BAYN BNP DTB3
## 2005-01-04 0.03702993 -1.8722247 -1.44514257 0.6495717 0.009090909
## 2005-01-05 -0.80868121 -2.8763703 -2.58036866 -0.5566617 0.009051383
## 2005-01-06 0.64429835 -0.1769240 0.04392226 1.4776002 0.008972332
## 2005-01-07 0.45643826 -0.4434562 -0.29134506 1.1845145 0.009051383
## 2005-01-10 -0.08729197 1.4130242 0.32976122 0.1810193 0.009209486
## 2005-01-11 -0.94221569 -1.8591429 -0.95377959 -0.3623669 0.009130435Extraemos los datos:
rESX<- rdata$EUROSTOX
rASML <- rdata$ASML
rBAYN <- rdata$BAYN
rBNP <- rdata$BNP
DTB3 <- rdata$DTB3Calculamos los “clean data” para los valores:
rESXc <- Return.clean(rESX, method = "boudt")
rASMLc <- Return.clean(rASML, method = "boudt")
rBAYNc <- Return.clean(rBAYN, method = "boudt")
rBNPc <- Return.clean(rBNP, method = "boudt")Veamos los retornos:
dygraph(rESX, main = "Rendimientos de índice EUROSTOX50") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(rASML, main = "Rendimientos de ASML Holding N.V.") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(rBAYN, main = "Rendimientos de Bayer AG") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(rBNP, main = "Rendimientos de BNP Paribas S.A.") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))3 Apartado 1
En primer lugar vamos a verificar si alguno de los retornos de los activos tiene alguna estructura:
fitASML <- auto.arima(rASML, trace = TRUE, test = "kpss", ic="bic")##
## ARIMA(2,0,2) with non-zero mean : 12354.17
## ARIMA(0,0,0) with non-zero mean : 12331.67
## ARIMA(1,0,0) with non-zero mean : 12337.19
## ARIMA(0,0,1) with non-zero mean : 12336.85
## ARIMA(0,0,0) with zero mean : 12325.78
## ARIMA(1,0,1) with non-zero mean : 12337.56
##
## Best model: ARIMA(0,0,0) with non-zero meanfitBAYN <- auto.arima(rBAYN, trace = TRUE, test = "kpss", ic="bic")##
## ARIMA(2,0,2) with non-zero mean : 11536.73
## ARIMA(0,0,0) with non-zero mean : 11524.08
## ARIMA(1,0,0) with non-zero mean : 11518.58
## ARIMA(0,0,1) with non-zero mean : 11517.48
## ARIMA(0,0,0) with zero mean : 11518.02
## ARIMA(1,0,1) with non-zero mean : 11521.54
## ARIMA(0,0,2) with non-zero mean : 11523.64
## ARIMA(1,0,2) with non-zero mean : 11528.65
## ARIMA(0,0,1) with zero mean : 11511.74
## ARIMA(1,0,1) with zero mean : 11516.29
## ARIMA(0,0,2) with zero mean : 11518.03
## ARIMA(1,0,2) with zero mean : 11523.57
##
## Best model: ARIMA(0,0,1) with zero meanfitBNP <- auto.arima(rBNP, trace = TRUE, test = "kpss", ic="bic")##
## ARIMA(2,0,2) with non-zero mean : 13446.54
## ARIMA(0,0,0) with non-zero mean : 13427.1
## ARIMA(1,0,0) with non-zero mean : 13435.52
## ARIMA(0,0,1) with non-zero mean : 13434.59
## ARIMA(0,0,0) with zero mean : 13419.2
## ARIMA(1,0,1) with non-zero mean : 13443.48
##
## Best model: ARIMA(0,0,0) with non-zero meanComo podemos verificar, todos los activos no poseen estructura ARIMA. El siguiente punto será verificar si existe el efecto de ARCH en los rendimientos. Para ello utilizaremos los residuos al cuadrado de de los modelos fit.
Vemamos el contraste de Ljung-Box:
Box.test(fitASML$residuals^2, lag=12, type = "Ljung-Box")##
## Box-Ljung test
##
## data: fitASML$residuals^2
## X-squared = 19.398, df = 12, p-value = 0.07938Box.test(fitBAYN$residuals^2, lag=12, type = "Ljung-Box")##
## Box-Ljung test
##
## data: fitBAYN$residuals^2
## X-squared = 63.662, df = 12, p-value = 4.814e-09Box.test(fitBNP$residuals^2, lag=12, type = "Ljung-Box")##
## Box-Ljung test
##
## data: fitBNP$residuals^2
## X-squared = 889.12, df = 12, p-value < 2.2e-16Recordemos que el contraste de Ljung-Box enfrenta la hipótesis nula de que no hay efectos arch frente a la alternativa de que existen efectos arch. Dado que los p-valores son inferiores al 5% rechazamos la hipótesis nula de la no existencia del efecto arch, por lo que podemos inturir que nos enferntamos a los efectos del arch.
Vamos a especificar el modelo GARCH(1,1) con el modelo de la media (0,0), ya que son los resultados que obtuvimos en el ajuste de ARIMA.
Veamos la especificación:
garch11.spec <- ugarchspec(variance.model = list(garchOrder=c(1,1)),
mean.model = list(armaOrder=c(0,0)))Especifiquemos los modelos arch de orden <=5 para cada activo:
arch.order <- 1:5
arch.names <- paste("arch", arch.order, sep=".")
arch.listASML <- list()
arch.listBAYN <- list()
arch.listBNP <- list()
for(p in arch.order) {
arch.spec <- ugarchspec(variance.model = list(garchOrder=c(p,0)),
mean.model = list(armaOrder=c(0,0)))
arch.fitASML <- ugarchfit(spec=arch.spec, data=rASML,
solver.control=list(trace = 0))
arch.listASML[[p]] <- arch.fitASML
arch.fitBAYN <- ugarchfit(spec=arch.spec, data=rBAYN,
solver.control=list(trace = 0))
arch.listBAYN[[p]] <- arch.fitBAYN
arch.fitBNP <- ugarchfit(spec=arch.spec, data=rBNP,
solver.control=list(trace = 0))
arch.listBNP[[p]] <- arch.fitBNP
}
names(arch.listASML) <- arch.names
info.matASML <- sapply(arch.listASML, infocriteria)
rownames(info.matASML) <- rownames(infocriteria(arch.listASML[[1]]))
info.matASML## arch.1 arch.2 arch.3 arch.4 arch.5
## Akaike 4.319460 4.311289 4.306464 4.289022 4.283014
## Bayes 4.325742 4.319664 4.316933 4.301585 4.297671
## Shibata 4.319458 4.311285 4.306458 4.289013 4.283002
## Hannan-Quinn 4.321726 4.314309 4.310240 4.293553 4.288300names(arch.listBAYN) <- arch.names
info.matBAYN <- sapply(arch.listBAYN, infocriteria)
rownames(info.matBAYN) <- rownames(infocriteria(arch.listBAYN[[1]]))
info.matBAYN## arch.1 arch.2 arch.3 arch.4 arch.5
## Akaike 4.001734 3.984056 3.972407 4.622346 4.444896
## Bayes 4.008015 3.992431 3.982876 4.634908 4.459553
## Shibata 4.001732 3.984052 3.972401 4.622337 4.444884
## Hannan-Quinn 4.004000 3.987077 3.976183 4.626877 4.450183names(arch.listBNP) <- arch.names
info.matBNP <- sapply(arch.listBNP, infocriteria)
rownames(info.matBNP) <- rownames(infocriteria(arch.listBNP[[1]]))
info.matBNP ## arch.1 arch.2 arch.3 arch.4 arch.5
## Akaike 4.574924 4.452698 4.407020 4.349775 4.322776
## Bayes 4.581205 4.461073 4.417489 4.362337 4.337432
## Shibata 4.574922 4.452694 4.407014 4.349766 4.322764
## Hannan-Quinn 4.577190 4.455719 4.410796 4.354306 4.328062Los criterios de información para: * ASML: mejoran al aumentar el orden del ARCH * BAYN: señalan como el mejor modelo el ARCH(3) * BNP: mejoran al aumentar el orden del ARCH
Ahora vamos a probar a ejecutar el modelo para cada activo:
3.1 ASML
Vamos a comparar los resultados de GARCH(1,1) con ARCH(5)
garch11.fitASML <- ugarchfit(spec = garch11.spec, data=rASML)
garch11.fitASML##
## *---------------------------------*
## * GARCH Model Fit *
## *---------------------------------*
##
## Conditional Variance Dynamics
## -----------------------------------
## GARCH Model : sGARCH(1,1)
## Mean Model : ARFIMA(0,0,0)
## Distribution : norm
##
## Optimal Parameters
## ------------------------------------
## Estimate Std. Error t value Pr(>|t|)
## mu 0.096673 0.037287 2.5927 0.009523
## omega 0.052327 0.007774 6.7308 0.000000
## alpha1 0.017665 0.002015 8.7664 0.000000
## beta1 0.970493 0.002130 455.5579 0.000000
##
## Robust Standard Errors:
## Estimate Std. Error t value Pr(>|t|)
## mu 0.096673 0.035350 2.7348 0.006242
## omega 0.052327 0.018122 2.8875 0.003883
## alpha1 0.017665 0.004289 4.1185 0.000038
## beta1 0.970493 0.001249 777.2193 0.000000
##
## LogLikelihood : -6069.38
##
## Information Criteria
## ------------------------------------
##
## Akaike 4.2725
## Bayes 4.2809
## Shibata 4.2725
## Hannan-Quinn 4.2755
##
## Weighted Ljung-Box Test on Standardized Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 1.350 0.2452
## Lag[2*(p+q)+(p+q)-1][2] 1.778 0.3025
## Lag[4*(p+q)+(p+q)-1][5] 4.240 0.2256
## d.o.f=0
## H0 : No serial correlation
##
## Weighted Ljung-Box Test on Standardized Squared Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 0.04575 0.8306
## Lag[2*(p+q)+(p+q)-1][5] 0.11204 0.9976
## Lag[4*(p+q)+(p+q)-1][9] 0.34787 0.9995
## d.o.f=2
##
## Weighted ARCH LM Tests
## ------------------------------------
## Statistic Shape Scale P-Value
## ARCH Lag[3] 0.003163 0.500 2.000 0.9551
## ARCH Lag[5] 0.073108 1.440 1.667 0.9915
## ARCH Lag[7] 0.230110 2.315 1.543 0.9960
##
## Nyblom stability test
## ------------------------------------
## Joint Statistic: 0.5661
## Individual Statistics:
## mu 0.03133
## omega 0.12554
## alpha1 0.16117
## beta1 0.16596
##
## Asymptotic Critical Values (10% 5% 1%)
## Joint Statistic: 1.07 1.24 1.6
## Individual Statistic: 0.35 0.47 0.75
##
## Sign Bias Test
## ------------------------------------
## t-value prob sig
## Sign Bias 0.3693 0.7120
## Negative Sign Bias 0.5875 0.5569
## Positive Sign Bias 1.2296 0.2190
## Joint Effect 1.8689 0.6001
##
##
## Adjusted Pearson Goodness-of-Fit Test:
## ------------------------------------
## group statistic p-value(g-1)
## 1 20 116.1 5.938e-16
## 2 30 132.6 3.371e-15
## 3 40 136.3 1.036e-12
## 4 50 161.4 6.426e-14
##
##
## Elapsed time : 0.249001vamos a probar la las estimación rolling en la que utilizaremos una ventana de 120 días con ajuste diario.
cl = makePSOCKcluster(10)
rollASML = ugarchroll(garch11.spec, rASML, n.start = 120, refit.every = 1,
refit.window = "moving", solver = "hybrid", calculate.VaR = TRUE,
VaR.alpha = c(0.01, 0.05), cluster = cl, keep.coef = TRUE)Veamos los resultados al 1%:
report(rollASML, type="VaR", VaR.alpha = 0.01, conf.level= 0.99)## VaR Backtest Report
## ===========================================
## Model: sGARCH-norm
## Backtest Length: 2723
## Data:
##
## ==========================================
## alpha: 1%
## Expected Exceed: 27.2
## Actual VaR Exceed: 48
## Actual %: 1.8%
##
## Unconditional Coverage (Kupiec)
## Null-Hypothesis: Correct Exceedances
## LR.uc Statistic: 13.041
## LR.uc Critical: 6.635
## LR.uc p-value: 0
## Reject Null: YES
##
## Conditional Coverage (Christoffersen)
## Null-Hypothesis: Correct Exceedances and
## Independence of Failures
## LR.cc Statistic: 13.068
## LR.cc Critical: 9.21
## LR.cc p-value: 0.001
## Reject Null: YESVeamos los resultados al 5%:
report(rollASML, type="VaR", VaR.alpha = 0.05, conf.level= 0.95)## VaR Backtest Report
## ===========================================
## Model: sGARCH-norm
## Backtest Length: 2723
## Data:
##
## ==========================================
## alpha: 5%
## Expected Exceed: 136.2
## Actual VaR Exceed: 128
## Actual %: 4.7%
##
## Unconditional Coverage (Kupiec)
## Null-Hypothesis: Correct Exceedances
## LR.uc Statistic: 0.524
## LR.uc Critical: 3.841
## LR.uc p-value: 0.469
## Reject Null: NO
##
## Conditional Coverage (Christoffersen)
## Null-Hypothesis: Correct Exceedances and
## Independence of Failures
## LR.cc Statistic: 1.18
## LR.cc Critical: 5.991
## LR.cc p-value: 0.554
## Reject Null: NOVeamos algunos gráficos:
plot(garch11.fitASML, which=9)
plot(garch11.fitASML, which=10)
plot(garch11.fitASML, which=11)
plot(garch11.fitASML, which=12)
Comparación entre GARCH(1,1) y ARCH(5):
par(mfrow=c(2,1))
plot.ts(sigma(garch11.fitASML), main="GARCH(1,1) conditional vol",
ylab="vol", col="blue")
plot.ts(sigma(arch.listASML$arch.5), main="ARCH(5) conditional vol",
ylab="vol", col="blue")
par(mfrow=c(1,1))En este caso podríamos elegir el GARCH(1,1) ya que sus criterios de información son mejores. Además, tiene menor número de parámetros lo que hace que el modelo sea más fiable a la hora de la explicación de los parámetros
3.2 BAYN
garch11.fitBAYN <- ugarchfit(spec = garch11.spec, data=rBAYN)
garch11.fitBAYN##
## *---------------------------------*
## * GARCH Model Fit *
## *---------------------------------*
##
## Conditional Variance Dynamics
## -----------------------------------
## GARCH Model : sGARCH(1,1)
## Mean Model : ARFIMA(0,0,0)
## Distribution : norm
##
## Optimal Parameters
## ------------------------------------
## Estimate Std. Error t value Pr(>|t|)
## mu 0.088435 0.029618 2.9858 0.002828
## omega 0.024979 0.006374 3.9191 0.000089
## alpha1 0.035715 0.002839 12.5813 0.000000
## beta1 0.957694 0.002595 369.1005 0.000000
##
## Robust Standard Errors:
## Estimate Std. Error t value Pr(>|t|)
## mu 0.088435 0.026485 3.3390 0.000841
## omega 0.024979 0.014624 1.7081 0.087616
## alpha1 0.035715 0.005905 6.0481 0.000000
## beta1 0.957694 0.005268 181.7864 0.000000
##
## LogLikelihood : -5539.376
##
## Information Criteria
## ------------------------------------
##
## Akaike 3.8997
## Bayes 3.9080
## Shibata 3.8997
## Hannan-Quinn 3.9027
##
## Weighted Ljung-Box Test on Standardized Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 9.846 0.001702
## Lag[2*(p+q)+(p+q)-1][2] 9.928 0.001996
## Lag[4*(p+q)+(p+q)-1][5] 10.885 0.005444
## d.o.f=0
## H0 : No serial correlation
##
## Weighted Ljung-Box Test on Standardized Squared Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 14.34 0.0001525
## Lag[2*(p+q)+(p+q)-1][5] 14.50 0.0006056
## Lag[4*(p+q)+(p+q)-1][9] 15.12 0.0033441
## d.o.f=2
##
## Weighted ARCH LM Tests
## ------------------------------------
## Statistic Shape Scale P-Value
## ARCH Lag[3] 0.06423 0.500 2.000 0.7999
## ARCH Lag[5] 0.17083 1.440 1.667 0.9722
## ARCH Lag[7] 0.86167 2.315 1.543 0.9350
##
## Nyblom stability test
## ------------------------------------
## Joint Statistic: 0.3533
## Individual Statistics:
## mu 0.04363
## omega 0.12801
## alpha1 0.11550
## beta1 0.09208
##
## Asymptotic Critical Values (10% 5% 1%)
## Joint Statistic: 1.07 1.24 1.6
## Individual Statistic: 0.35 0.47 0.75
##
## Sign Bias Test
## ------------------------------------
## t-value prob sig
## Sign Bias 1.5320 0.1256
## Negative Sign Bias 0.6513 0.5149
## Positive Sign Bias 1.1752 0.2400
## Joint Effect 4.9179 0.1779
##
##
## Adjusted Pearson Goodness-of-Fit Test:
## ------------------------------------
## group statistic p-value(g-1)
## 1 20 137.8 4.711e-20
## 2 30 163.7 9.736e-21
## 3 40 184.4 9.062e-21
## 4 50 180.6 5.926e-17
##
##
## Elapsed time : 0.2017839vamos a probar la las estimación rolling en la que utilizaremos una ventana de 120 días con ajuste diario.
cl = makePSOCKcluster(10)
rollBAYN = ugarchroll(garch11.spec, rBAYN, n.start = 120, refit.every = 1,
refit.window = "moving", solver = "hybrid", calculate.VaR = TRUE,
VaR.alpha = c(0.01, 0.05), cluster = cl, keep.coef = TRUE)## Warning: cerrando la conenexion 14 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 13 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 12 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 11 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 10 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 9 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 8 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 7 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 6 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 5 (<-Wolfshund:11969) que no esta siendo
## utilizadaVeamos los resultados al 1%:
report(rollBAYN, type="VaR", VaR.alpha = 0.01, conf.level= 0.99)## VaR Backtest Report
## ===========================================
## Model: sGARCH-norm
## Backtest Length: 2723
## Data:
##
## ==========================================
## alpha: 1%
## Expected Exceed: 27.2
## Actual VaR Exceed: 53
## Actual %: 1.9%
##
## Unconditional Coverage (Kupiec)
## Null-Hypothesis: Correct Exceedances
## LR.uc Statistic: 19.3
## LR.uc Critical: 6.635
## LR.uc p-value: 0
## Reject Null: YES
##
## Conditional Coverage (Christoffersen)
## Null-Hypothesis: Correct Exceedances and
## Independence of Failures
## LR.cc Statistic: 19.301
## LR.cc Critical: 9.21
## LR.cc p-value: 0
## Reject Null: YESVeamos los resultados al 5%:
report(rollBAYN, type="VaR", VaR.alpha = 0.05, conf.level= 0.95)## VaR Backtest Report
## ===========================================
## Model: sGARCH-norm
## Backtest Length: 2723
## Data:
##
## ==========================================
## alpha: 5%
## Expected Exceed: 136.2
## Actual VaR Exceed: 155
## Actual %: 5.7%
##
## Unconditional Coverage (Kupiec)
## Null-Hypothesis: Correct Exceedances
## LR.uc Statistic: 2.635
## LR.uc Critical: 3.841
## LR.uc p-value: 0.105
## Reject Null: NO
##
## Conditional Coverage (Christoffersen)
## Null-Hypothesis: Correct Exceedances and
## Independence of Failures
## LR.cc Statistic: 3.802
## LR.cc Critical: 5.991
## LR.cc p-value: 0.149
## Reject Null: NOVeamos algunos gráficos:
plot(garch11.fitASML, which=9)
plot(garch11.fitASML, which=10)
plot(garch11.fitASML, which=11)
plot(garch11.fitASML, which=12)
Comparación entre GARCH(1,1) y ARCH(3):
par(mfrow=c(2,1))
plot.ts(sigma(garch11.fitBAYN), main="GARCH(1,1) conditional vol",
ylab="vol", col="blue")
plot.ts(sigma(arch.listBAYN$arch.3), main="ARCH(3) conditional vol",
ylab="vol", col="blue")
par(mfrow=c(1,1))En este caso al igual que en el anterior los criterios de información señalan el modelo GARCH(1,1) como mejor.
3.3 BNP
garch11.fitBNP <- ugarchfit(spec = garch11.spec, data=rBNP)
garch11.fitBNP##
## *---------------------------------*
## * GARCH Model Fit *
## *---------------------------------*
##
## Conditional Variance Dynamics
## -----------------------------------
## GARCH Model : sGARCH(1,1)
## Mean Model : ARFIMA(0,0,0)
## Distribution : norm
##
## Optimal Parameters
## ------------------------------------
## Estimate Std. Error t value Pr(>|t|)
## mu 0.065288 0.031787 2.0539 0.039987
## omega 0.054400 0.015927 3.4155 0.000637
## alpha1 0.101417 0.012365 8.2017 0.000000
## beta1 0.893822 0.012696 70.4009 0.000000
##
## Robust Standard Errors:
## Estimate Std. Error t value Pr(>|t|)
## mu 0.065288 0.029197 2.2361 0.025346
## omega 0.054400 0.020847 2.6095 0.009069
## alpha1 0.101417 0.022644 4.4787 0.000008
## beta1 0.893822 0.021426 41.7158 0.000000
##
## LogLikelihood : -6071.469
##
## Information Criteria
## ------------------------------------
##
## Akaike 4.2740
## Bayes 4.2824
## Shibata 4.2740
## Hannan-Quinn 4.2770
##
## Weighted Ljung-Box Test on Standardized Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 0.06101 0.80490
## Lag[2*(p+q)+(p+q)-1][2] 1.04156 0.48495
## Lag[4*(p+q)+(p+q)-1][5] 7.32460 0.04323
## d.o.f=0
## H0 : No serial correlation
##
## Weighted Ljung-Box Test on Standardized Squared Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 0.1005 0.7512
## Lag[2*(p+q)+(p+q)-1][5] 0.6038 0.9405
## Lag[4*(p+q)+(p+q)-1][9] 1.6258 0.9447
## d.o.f=2
##
## Weighted ARCH LM Tests
## ------------------------------------
## Statistic Shape Scale P-Value
## ARCH Lag[3] 0.1223 0.500 2.000 0.7265
## ARCH Lag[5] 0.9359 1.440 1.667 0.7521
## ARCH Lag[7] 1.7461 2.315 1.543 0.7708
##
## Nyblom stability test
## ------------------------------------
## Joint Statistic: 1.0693
## Individual Statistics:
## mu 0.05103
## omega 0.63217
## alpha1 0.15334
## beta1 0.23540
##
## Asymptotic Critical Values (10% 5% 1%)
## Joint Statistic: 1.07 1.24 1.6
## Individual Statistic: 0.35 0.47 0.75
##
## Sign Bias Test
## ------------------------------------
## t-value prob sig
## Sign Bias 0.1285 0.8978
## Negative Sign Bias 0.8903 0.3734
## Positive Sign Bias 1.0459 0.2957
## Joint Effect 3.0170 0.3890
##
##
## Adjusted Pearson Goodness-of-Fit Test:
## ------------------------------------
## group statistic p-value(g-1)
## 1 20 46.45 4.282e-04
## 2 30 68.71 4.544e-05
## 3 40 72.33 9.322e-04
## 4 50 91.77 2.069e-04
##
##
## Elapsed time : 0.1530271vamos a probar la las estimación rolling en la que utilizaremos una ventana de 120 días con ajuste diario.
cl = makePSOCKcluster(10)
rollBNP = ugarchroll(garch11.spec, rBNP, n.start = 120, refit.every = 1,
refit.window = "moving", solver = "hybrid", calculate.VaR = TRUE,
VaR.alpha = c(0.01, 0.05), cluster = cl, keep.coef = TRUE)Veamos los resultados al 1%:
report(rollBNP, type="VaR", VaR.alpha = 0.01, conf.level= 0.99)## VaR Backtest Report
## ===========================================
## Model: sGARCH-norm
## Backtest Length: 2723
## Data:
##
## ==========================================
## alpha: 1%
## Expected Exceed: 27.2
## Actual VaR Exceed: 57
## Actual %: 2.1%
##
## Unconditional Coverage (Kupiec)
## Null-Hypothesis: Correct Exceedances
## LR.uc Statistic: 25.005
## LR.uc Critical: 6.635
## LR.uc p-value: 0
## Reject Null: YES
##
## Conditional Coverage (Christoffersen)
## Null-Hypothesis: Correct Exceedances and
## Independence of Failures
## LR.cc Statistic: 25.481
## LR.cc Critical: 9.21
## LR.cc p-value: 0
## Reject Null: YESVeamos los resultados al 5%:
report(rollBNP, type="VaR", VaR.alpha = 0.05, conf.level= 0.95)## VaR Backtest Report
## ===========================================
## Model: sGARCH-norm
## Backtest Length: 2723
## Data:
##
## ==========================================
## alpha: 5%
## Expected Exceed: 136.2
## Actual VaR Exceed: 158
## Actual %: 5.8%
##
## Unconditional Coverage (Kupiec)
## Null-Hypothesis: Correct Exceedances
## LR.uc Statistic: 3.518
## LR.uc Critical: 3.841
## LR.uc p-value: 0.061
## Reject Null: NO
##
## Conditional Coverage (Christoffersen)
## Null-Hypothesis: Correct Exceedances and
## Independence of Failures
## LR.cc Statistic: 3.907
## LR.cc Critical: 5.991
## LR.cc p-value: 0.142
## Reject Null: NOVeamos algunos gráficos:
plot(garch11.fitASML, which=9)
plot(garch11.fitASML, which=10)
plot(garch11.fitASML, which=11)
plot(garch11.fitASML, which=12)
Vamos a comparar ARCH(5) con GARCH(1,1)
par(mfrow=c(2,1))
plot.ts(sigma(garch11.fitBNP), main="GARCH(1,1) conditional vol",
ylab="vol", col="blue")
plot.ts(sigma(arch.listBNP$arch.5), main="ARCH(5) conditional vol",
ylab="vol", col="blue")
par(mfrow=c(1,1))En este caso, al igual que en los anteriores, el modelo GARCH(1,1) tiene mejores criterios de información, pero hay que fijarse que la volatilidad condicional se parece bastante entre los dos modeos.
4 Apartado 2: Ajuste de los modelos CAPM para los activos
En este apartado vamos a ajustar tres modelos distintos de CAPM para cada activos. Son los siguientes modelos:
$(r - R_F) = (R_M - R_F) + $
$(r - R_F) = (R_M - R_F) + $
\(r = \alpha + \beta R_F + \varepsilon\)
Vamos a calcular los tipos reales de las rentabilidades y de mercado:
EX.ASML <- rASML-DTB3
EX.BAYN <- rBAYN-DTB3
EX.BNP <- rBNP-DTB3
EX.MKT <- rESX-DTB34.1 Modelos CAPM para ASML
Vamos a proceder el cálculo de los moodelos para ASML.
4.1.1 Modelo 1
m1.ASML <- lm(EX.ASML ~ EX.MKT - 1)
sm1.ASML <- summary(m1.ASML)
print(sm1.ASML)##
## Call:
## lm(formula = EX.ASML ~ EX.MKT - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -22.8700 -0.8441 0.0230 0.9380 20.4790
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## EX.MKT 0.7871 0.0236 33.35 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.79 on 2842 degrees of freedom
## Multiple R-squared: 0.2813, Adjusted R-squared: 0.2811
## F-statistic: 1112 on 1 and 2842 DF, p-value: < 2.2e-16print(m1.ASML$coefficients)## EX.MKT
## 0.7871427m1.ASML.coef <- m1.ASML$coefficients
beta1.ASML <- m1.ASML.coef
cat("Beta ", beta1.ASML, "\n")## Beta 0.7871427Comprobamos los resíduos del modelo:
uhat1.ASML <- m1.ASML$residuals
uhat1.ASML <- as.xts(uhat1.ASML)
ggAcf(uhat1.ASML) + labs(title="Residuos CAPM1")
uhat21.ASML <- uhat1.ASML^2
ggAcf(uhat21.ASML) + labs(title="Residuos al cuadrado de CAPM1")
Box.test(uhat21.ASML, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat21.ASML
## X-squared = 3.3037, df = 12, p-value = 0.993ArchTest(uhat1.ASML)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat1.ASML
## Chi-squared = 3.1623, df = 12, p-value = 0.99434.1.2 Modelo 2
m2.ASML <- lm(EX.ASML ~ EX.MKT)
sm2.ASML <- summary(m2.ASML)
print(sm2.ASML)##
## Call:
## lm(formula = EX.ASML ~ EX.MKT)
##
## Residuals:
## Min 1Q Median 3Q Max
## -22.9354 -0.9098 -0.0434 0.8718 20.4130
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.06569 0.03356 1.957 0.0504 .
## EX.MKT 0.78770 0.02359 33.392 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.789 on 2841 degrees of freedom
## Multiple R-squared: 0.2819, Adjusted R-squared: 0.2816
## F-statistic: 1115 on 1 and 2841 DF, p-value: < 2.2e-16print(m2.ASML$coefficients)## (Intercept) EX.MKT
## 0.06568963 0.78770490m2.ASML.coef <- m2.ASML$coefficients
alpha2.ASML <- m2.ASML.coef[1]
beta2.ASML <- m2.ASML.coef[2]
cat("Alpha ", alpha2.ASML, "\n")## Alpha 0.06568963cat("Beta ", beta2.ASML, "\n")## Beta 0.7877049Comprobamos los resíduos del modelo:
uhat2.ASML <- m2.ASML$residuals
uhat2.ASML <- as.xts(uhat2.ASML)
ggAcf(uhat2.ASML) + labs(title="Residuos CAPM2")
uhat22.ASML <- uhat2.ASML^2
ggAcf(uhat22.ASML) + labs(title="Residuos al cuadrado de CAPM2")
Box.test(uhat22.ASML, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat22.ASML
## X-squared = 3.3444, df = 12, p-value = 0.9926ArchTest(uhat2.ASML)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat2.ASML
## Chi-squared = 3.2003, df = 12, p-value = 0.9944.1.3 Modelo 3
m3.ASML <- lm(rASML ~ rESX)
sm3.ASML<- summary(m3.ASML)
print(sm3.ASML)##
## Call:
## lm(formula = rASML ~ rESX)
##
## Residuals:
## Min 1Q Median 3Q Max
## -22.9363 -0.9094 -0.0435 0.8729 20.4135
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.06670 0.03356 1.988 0.047 *
## rESX 0.78768 0.02359 33.393 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.789 on 2841 degrees of freedom
## Multiple R-squared: 0.2819, Adjusted R-squared: 0.2816
## F-statistic: 1115 on 1 and 2841 DF, p-value: < 2.2e-16m3.ASML.coef <- m3.ASML$coefficients
alpha3.ASML <- m3.ASML.coef[1]
beta3.ASML <- m3.ASML.coef[2]
cat("Alpha ",alpha3.ASML, "\n")## Alpha 0.06670355cat("Beta ", beta3.ASML, "\n")## Beta 0.7876765Comprobamos los resíduos del modelo:
uhat3.ASML <- m3.ASML$residuals
uhat3.ASML <- as.xts(uhat3.ASML)
ggAcf(uhat3.ASML) + labs(title="Residuos de CAPM3")
uhat23.ASML <- uhat3.ASML^2
ggAcf(uhat23.ASML) + labs(title="Residuos al cuadrado de CAPM3")
Box.test(uhat23.ASML, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat23.ASML
## X-squared = 3.3437, df = 12, p-value = 0.9926ArchTest(uhat3.ASML)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat3.ASML
## Chi-squared = 3.1998, df = 12, p-value = 0.994Veamos los criterios de información para ASML
AIC.ASML <- AIC(m1.ASML,m2.ASML,m3.ASML)
BIC.ASML <- BIC(m1.ASML,m2.ASML,m3.ASML)
info.capm.ASML <- cbind(AIC.ASML,BIC.ASML)
info.capm.ASML <- info.capm.ASML[,-c(1,3)]
rownames(info.capm.ASML) <- c("CAPM1", "CAPM2", "CAPM3")
colnames(info.capm.ASML) <- c("Akaike", "Schwarz")
info.capm.ASML## Akaike Schwarz
## CAPM1 11382.32 11394.22
## CAPM2 11380.49 11398.34
## CAPM3 11380.46 11398.31Dados los criterios de información y que el parámetro alpha es significativo se puede seleccionar el modelo CAPM3
4.2 Modelos CAPM para BAYN
Vamos a proceder el cálculo de los moodelos para BAYN.
4.2.1 Modelo 1
m1.BAYN <- lm(EX.BAYN ~ EX.MKT - 1)
sm1.BAYN <- summary(m1.BAYN)
print(sm1.BAYN)##
## Call:
## lm(formula = EX.BAYN ~ EX.MKT - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -14.1472 -0.6158 0.0166 0.6725 25.8410
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## EX.MKT 0.85975 0.01798 47.83 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.364 on 2842 degrees of freedom
## Multiple R-squared: 0.446, Adjusted R-squared: 0.4458
## F-statistic: 2288 on 1 and 2842 DF, p-value: < 2.2e-16print(m1.BAYN$coefficients)## EX.MKT
## 0.8597525m1.BAYN.coef <- m1.BAYN$coefficients
beta1.BAYN <- m1.BAYN.coef
cat("Beta ", beta1.BAYN, "\n")## Beta 0.8597525Comprobamos los resíduos del modelo:
uhat1.BAYN <- m1.BAYN$residuals
uhat1.BAYN <- as.xts(uhat1.BAYN)
ggAcf(uhat1.BAYN) + labs(title="Residuos CAPM1")
uhat21.BAYN <- uhat1.BAYN^2
ggAcf(uhat21.BAYN) + labs(title="Residuos al cuadrado de CAPM1")
Box.test(uhat21.BAYN, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat21.BAYN
## X-squared = 23.266, df = 12, p-value = 0.02555ArchTest(uhat1.BAYN)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat1.BAYN
## Chi-squared = 22.77, df = 12, p-value = 0.029754.2.2 Modelo 2
m2.BAYN <- lm(EX.BAYN ~ EX.MKT)
sm2.BAYN <- summary(m2.BAYN)
print(sm2.BAYN)##
## Call:
## lm(formula = EX.BAYN ~ EX.MKT)
##
## Residuals:
## Min 1Q Median 3Q Max
## -14.2051 -0.6733 -0.0408 0.6149 25.7838
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.05737 0.02556 2.245 0.0249 *
## EX.MKT 0.86024 0.01796 47.886 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.363 on 2841 degrees of freedom
## Multiple R-squared: 0.4466, Adjusted R-squared: 0.4464
## F-statistic: 2293 on 1 and 2841 DF, p-value: < 2.2e-16print(m2.BAYN$coefficients)## (Intercept) EX.MKT
## 0.05737158 0.86024351m2.BAYN.coef <- m2.BAYN$coefficients
alpha2.BAYN <- m2.BAYN.coef[1]
beta2.BAYN <- m2.BAYN.coef[2]
cat("Alpha ", alpha2.BAYN, "\n")## Alpha 0.05737158cat("Beta ", beta2.BAYN, "\n")## Beta 0.8602435Comprobamos los resíduos del modelo:
uhat2.BAYN <- m2.BAYN$residuals
uhat2.BAYN <- as.xts(uhat2.BAYN)
ggAcf(uhat2.BAYN) + labs(title="Residuos CAPM2")
uhat22.BAYN <- uhat2.BAYN^2
ggAcf(uhat22.BAYN) + labs(title="Residuos al cuadrado de CAPM2")
Box.test(uhat22.BAYN, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat22.BAYN
## X-squared = 23.817, df = 12, p-value = 0.02154ArchTest(uhat2.BAYN)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat2.BAYN
## Chi-squared = 23.286, df = 12, p-value = 0.025394.2.3 Modelo 3
m3.BAYN <- lm(rBAYN ~ rESX)
sm3.BAYN<- summary(m3.BAYN)
print(sm3.BAYN)##
## Call:
## lm(formula = rBAYN ~ rESX)
##
## Residuals:
## Min 1Q Median 3Q Max
## -14.2057 -0.6724 -0.0414 0.6143 25.7833
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.05804 0.02556 2.271 0.0232 *
## rESX 0.86030 0.01796 47.891 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.363 on 2841 degrees of freedom
## Multiple R-squared: 0.4467, Adjusted R-squared: 0.4465
## F-statistic: 2294 on 1 and 2841 DF, p-value: < 2.2e-16m3.BAYN.coef <- m3.BAYN$coefficients
alpha3.BAYN <- m3.BAYN.coef[1]
beta3.BAYN <- m3.BAYN.coef[2]
cat("Alpha ",alpha3.BAYN, "\n")## Alpha 0.05803995cat("Beta ", beta3.BAYN, "\n")## Beta 0.860296Comprobamos los resíduos del modelo:
uhat3.BAYN <- m3.BAYN$residuals
uhat3.BAYN <- as.xts(uhat3.BAYN)
ggAcf(uhat3.BAYN) + labs(title="Residuos de CAPM3")
uhat23.BAYN <- uhat3.BAYN^2
ggAcf(uhat23.BAYN) + labs(title="Residuos al cuadrado de CAPM3")
Box.test(uhat23.BAYN, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat23.BAYN
## X-squared = 23.823, df = 12, p-value = 0.0215ArchTest(uhat3.BAYN)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat3.BAYN
## Chi-squared = 23.291, df = 12, p-value = 0.02535Veamos los criterios de información para BAYN
AIC.BAYN <- AIC(m1.BAYN,m2.BAYN,m3.BAYN)
BIC.BAYN <- BIC(m1.BAYN,m2.BAYN,m3.BAYN)
info.capm.BAYN <- cbind(AIC.BAYN,BIC.BAYN)
info.capm.BAYN <- info.capm.BAYN[,-c(1,3)]
rownames(info.capm.BAYN) <- c("CAPM1", "CAPM2", "CAPM3")
colnames(info.capm.BAYN) <- c("Akaike", "Schwarz")
info.capm.BAYN## Akaike Schwarz
## CAPM1 9834.585 9846.490
## CAPM2 9831.547 9849.405
## CAPM3 9831.586 9849.444En este caso el parámetro alpha también es significativo, y nos ijaremos el el parmámetro de Akaike y, por tanto, podemos seleccionar el modelo CAMP2.
4.3 Modelos CAPM para BNP
Vamos a proceder el cálculo de los moodelos para BNP.
4.3.1 Modelo 1
m1.BNP <- lm(EX.BNP ~ EX.MKT - 1)
sm1.BNP <- summary(m1.BNP)
print(sm1.BNP)##
## Call:
## lm(formula = EX.BNP ~ EX.MKT - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.5833 -0.6100 -0.0065 0.6661 12.7391
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## EX.MKT 1.42108 0.02069 68.7 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.569 on 2842 degrees of freedom
## Multiple R-squared: 0.6241, Adjusted R-squared: 0.624
## F-statistic: 4719 on 1 and 2842 DF, p-value: < 2.2e-16print(m1.BNP$coefficients)## EX.MKT
## 1.421083m1.BNP.coef <- m1.BNP$coefficients
beta1.BNP <- m1.BNP.coef
cat("Beta ", beta1.BNP, "\n")## Beta 1.421083Comprobamos los resíduos del modelo:
uhat1.BNP <- m1.BNP$residuals
uhat1.BNP <- as.xts(uhat1.BNP)
ggAcf(uhat1.BNP) + labs(title="Residuos CAPM1")## Warning: cerrando la conenexion 24 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 23 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 22 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 21 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 20 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 19 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 18 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 17 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 16 (<-Wolfshund:11969) que no esta siendo
## utilizada## Warning: cerrando la conenexion 15 (<-Wolfshund:11969) que no esta siendo
## utilizada
uhat21.BNP <- uhat1.BNP^2
ggAcf(uhat21.BNP) + labs(title="Residuos al cuadrado de CAPM1")
Box.test(uhat21.BNP, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat21.BNP
## X-squared = 850.29, df = 12, p-value < 2.2e-16ArchTest(uhat1.BNP)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat1.BNP
## Chi-squared = 386.47, df = 12, p-value < 2.2e-164.3.2 Modelo 2
m2.BNP <- lm(EX.BNP ~ EX.MKT)
sm2.BNP <- summary(m2.BNP)
print(sm2.BNP)##
## Call:
## lm(formula = EX.BNP ~ EX.MKT)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.6142 -0.6410 -0.0377 0.6354 12.7070
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.03096 0.02943 1.052 0.293
## EX.MKT 1.42135 0.02069 68.707 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.569 on 2841 degrees of freedom
## Multiple R-squared: 0.6243, Adjusted R-squared: 0.6242
## F-statistic: 4721 on 1 and 2841 DF, p-value: < 2.2e-16print(m2.BNP$coefficients)## (Intercept) EX.MKT
## 0.03096202 1.42134778m2.BNP.coef <- m2.BNP$coefficients
alpha2.BNP <- m2.BNP.coef[1]
beta2.BNP <- m2.BNP.coef[2]
cat("Alpha ", alpha2.BNP, "\n")## Alpha 0.03096202cat("Beta ", beta2.BNP, "\n")## Beta 1.421348Comprobamos los resíduos del modelo:
uhat2.BNP <- m2.BNP$residuals
uhat2.BNP <- as.xts(uhat2.BNP)
ggAcf(uhat2.BNP) + labs(title="Residuos CAPM2")
uhat22.BNP <- uhat2.BNP^2
ggAcf(uhat22.BNP) + labs(title="Residuos al cuadrado de CAPM2")
Box.test(uhat22.BNP, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat22.BNP
## X-squared = 847.95, df = 12, p-value < 2.2e-16ArchTest(uhat2.BNP)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat2.BNP
## Chi-squared = 386.48, df = 12, p-value < 2.2e-164.3.3 Modelo 3
m3.BNP <- lm(rBNP ~ rESX)
sm3.BNP<- summary(m3.BNP)
print(sm3.BNP)##
## Call:
## lm(formula = rBNP ~ rESX)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.6123 -0.6401 -0.0397 0.6359 12.7087
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.02895 0.02943 0.984 0.325
## rESX 1.42133 0.02069 68.710 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.569 on 2841 degrees of freedom
## Multiple R-squared: 0.6243, Adjusted R-squared: 0.6242
## F-statistic: 4721 on 1 and 2841 DF, p-value: < 2.2e-16m3.BNP.coef <- m3.BNP$coefficients
alpha3.BNP <- m3.BNP.coef[1]
beta3.BNP <- m3.BNP.coef[2]
cat("Alpha ",alpha3.BNP, "\n")## Alpha 0.02894877cat("Beta ", beta3.BNP, "\n")## Beta 1.421333Comprobamos los resíduos del modelo:
uhat3.BNP <- m3.BNP$residuals
uhat3.BNP <- as.xts(uhat3.BNP)
ggAcf(uhat3.BNP) + labs(title="Residuos de CAPM3")
uhat23.BNP <- uhat3.BNP^2
ggAcf(uhat23.BNP) + labs(title="Residuos al cuadrado de CAPM3")
Box.test(uhat23.BNP, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat23.BNP
## X-squared = 848.03, df = 12, p-value < 2.2e-16ArchTest(uhat3.BNP)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat3.BNP
## Chi-squared = 386.47, df = 12, p-value < 2.2e-16Veamos los criterios de información para BNP
AIC.BNP <- AIC(m1.BNP,m2.BNP,m3.BNP)
BIC.BNP <- BIC(m1.BNP,m2.BNP,m3.BNP)
info.capm.BNP <- cbind(AIC.BNP,BIC.BNP)
info.capm.BNP <- info.capm.BNP[,-c(1,3)]
rownames(info.capm.BNP) <- c("CAPM1", "CAPM2", "CAPM3")
colnames(info.capm.BNP) <- c("Akaike", "Schwarz")
info.capm.BNP## Akaike Schwarz
## CAPM1 10633.07 10644.97
## CAPM2 10633.96 10651.82
## CAPM3 10633.95 10651.80En este caso el parámetro de aplha no es significativo y los crtiterios de información señalan el modelo CAPM1 por lo que seleccionamos ese modelo.
4.4 Índices
En este apartado vamos a calcular tres índices: Sharpe, Treynor y Jensen, con el objetivo de ordenar los activos según su comportamiento.
4.4.1 índice de Sharpe
El Ratio de Sharpe se define como:
\[ Sharpe~Ratio = \frac{R_{Act}-R_{sin~riesgo}}{\sigma_{Act}}\]
Para simplificar suponemos que \(R_{sin~riesgo}=0\).
sharpe.ratio.ASML <- SharpeRatio(EX.ASML, FUN = "StdDev")
sharpe.ratio.BAYN <- SharpeRatio(EX.BAYN, FUN = "StdDev")
sharpe.ratio.BNP <- SharpeRatio(EX.BNP, FUN = "StdDev")
sharpe.ratio.ASML## ASML
## StdDev Sharpe (Rf=0%, p=95%): 0.02465285sharpe.ratio.BAYN## BAYN
## StdDev Sharpe (Rf=0%, p=95%): 0.02318896sharpe.ratio.BNP## BNP
## StdDev Sharpe (Rf=0%, p=95%): 0.0024775284.4.2 índice de Treynor
El Ratio de Treynor se define como:
\[ Treynor~Ratio = \frac{R_{Act}-R_{sin~riesgo}}{\beta_{Act}}\].
Para simplificar suponemos que \(R_{sin~riesgo}=0\)
treynor.ratio.ASML <- mean(EX.ASML)/ m2.ASML$coeff[2]
treynor.ratio.BAYN <- mean(EX.BAYN)/ m2.BAYN$coeff[2]
treynor.ratio.BNP <- mean(EX.BNP)/ m2.BNP$coeff[2]
treynor<- cbind(treynor.ratio.ASML,treynor.ratio.BAYN,treynor.ratio.BNP)
colnames(treynor) <- c("ASML", "BAYN", "BNP")
rownames(treynor) <- c("Treynor Ratio")
treynor## ASML BAYN BNP
## Treynor Ratio 0.06607171 0.04937026 0.0044615774.4.3 índice de Jensen
El alpha de Jensen se define como:
\[ Alpha~de~Jensen = \alpha=(R_{Act}-R_{sin~riesgo})-(R_{Act}-R_{sin~riesgo})\beta_{Act}\] E alpha de Jensen equivale al valor de alpha que estimamos en nuestros modelos CAPM.
jensen.ratio.ASML <- alpha2.ASML
jensen.ratio.BAYN <- alpha2.BAYN
jensen.ratio.BNP <- alpha2.BNP
jensen <- cbind(jensen.ratio.ASML, jensen.ratio.BAYN,jensen.ratio.BNP)
colnames(jensen) <- c("ASML", "BAYN", "BNP")
rownames(jensen) <- c("Alpha")
jensen## ASML BAYN BNP
## Alpha 0.06568963 0.05737158 0.030962024.5 Comportamiento de los activos
Los activos con \(\beta < 1\) se consideran defensivos, ya que tienen menor variabilidad que la del mercado. Los fondos con \(\beta = 1\) se consideran neutros, ya que sus rendimientos se comportan como el mercado. Los fondos con \(\beta > 1\) se consideran agresivos, ya que tienen mayor variabilidad que la del mercado. Los fondos con \(\beta = 1\) se consideran neutros, ya que sus rendimientos se comportan como el mercado.
veamos la tabla con las betas de los activos para ver su comportamiento de acuerdo al mercado:
beta.activos <- cbind(beta1.ASML, beta1.BAYN, beta1.BNP)
colnames(beta.activos) <- c("ASML", "BAYN", "BNP")
rownames(beta.activos) <- c("Beta")
beta.activos## ASML BAYN BNP
## Beta 0.7871427 0.8597525 1.421083Viendo como son los parámetros \(\beta\) ASML es el más defesivo seguido de BAYN que también lo es, pero en menor medida. El activo BNP es el que se comprta de forma agresiva.
5 Apartado 3: los parámetros de los CAPM y los índices son constantes en el tiempo?
En este apartado vamos a probar si los distintos parámetros elegidos son constantes o no en el tiempo.
Para conseguirlo vamos a probar dos maneras de demostrarlo: 1. Recursive. Vasmos a calcular la correlación recursiva de los parámetros fuiando como el comienzo en 100 2.Rolling. Vamos a cambiar del punto de vista fijando pantallas de 100 observaciones y moviendola día a día.
Este método lo aplicarmos a los CAPM2 para ver el comportamiento de los dos parámetros: alpha y beta, con el fin de tener una muestra homogénea.
5.1 ASML
Recursive:
ASML.matrix <- cbind(EX.ASML, EX.MKT)
ASML.matrix <- coredata(ASML.matrix)
ASML.X <- ASML.matrix[,2]
ASML.Y <- ASML.matrix[,1]
k0 <- 99
T <- length(ASML.X)
alpha.ASML <- beta.ASML <- array(dim=(T-k0))
for(i in 1:(T-k0))
{
k <- k0 + i
vX.ASML <- ASML.X[1:k]
vY.ASML <- ASML.Y[1:k]
m2.ASML <- lm(vY.ASML ~ vX.ASML)
alpha.ASML[i] <-m2.ASML$coeff[1]
beta.ASML[i] <- m2.ASML$coeff[2]
}
alpha.ASML <- xts(alpha.ASML, index(EX.ASML)[(k0+1):(T)])
beta.ASML <- xts(beta.ASML, index(EX.ASML)[(k0+1):(T)])
dygraph(alpha.ASML, main = "Recursive Alpha: ASML") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(beta.ASML, main = "Recursive Beta: ASML") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Rolling:
N <- 100
alpha.ASML <- beta.ASML <- array(dim=(T-N+1))
for(i in 1:(T-N+1))
{
vX.ASML <- ASML.X[i:(i+N)]
vY.ASML <- ASML.Y[i:(i+N)]
m2.ASML <- lm(vY.ASML ~ vX.ASML)
alpha.ASML[i] <-m2.ASML$coeff[1]
beta.ASML[i] <- m2.ASML$coeff[2]
}
alpha.ASML <- xts(alpha.ASML, index(EX.ASML)[(k0+1):(T)])
beta.ASML <- xts(beta.ASML, index(EX.ASML)[(k0+1):(T)])
dygraph(alpha.ASML, main = "Rolling Alpha: ASML") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(beta.ASML, main = "Rolling Beta: ASML") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))5.2 BAYN
Recursive:
BAYN.matrix <- cbind(EX.BAYN, EX.MKT)
BAYN.matrix <- coredata(BAYN.matrix)
BAYN.X <- BAYN.matrix[,2]
BAYN.Y <- BAYN.matrix[,1]
k0 <- 99
T <- length(BAYN.X)
alpha.BAYN <- beta.BAYN <- array(dim=(T-k0))
for(i in 1:(T-k0))
{
k <- k0 + i
vX.BAYN <- BAYN.X[1:k]
vY.BAYN <- BAYN.Y[1:k]
m2.BAYN <- lm(vY.BAYN ~ vX.BAYN)
alpha.BAYN[i] <-m2.BAYN$coeff[1]
beta.BAYN[i] <- m2.BAYN$coeff[2]
}
alpha.BAYN <- xts(alpha.BAYN, index(EX.BAYN)[(k0+1):(T)])
beta.BAYN <- xts(beta.BAYN, index(EX.BAYN)[(k0+1):(T)])
dygraph(alpha.BAYN, main = "Recursive Alpha: BAYN") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(beta.BAYN, main = "Recursive Beta: BAYN") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Rolling:
N <- 100
alpha.BAYN <- beta.BAYN <- array(dim=(T-N+1))
for(i in 1:(T-N+1))
{
vX.BAYN <- BAYN.X[i:(i+N)]
vY.BAYN <- BAYN.Y[i:(i+N)]
m2.BAYN <- lm(vY.BAYN ~ vX.BAYN)
alpha.BAYN[i] <-m2.BAYN$coeff[1]
beta.BAYN[i] <- m2.BAYN$coeff[2]
}
alpha.BAYN <- xts(alpha.BAYN, index(EX.BAYN)[(k0+1):(T)])
beta.BAYN <- xts(beta.BAYN, index(EX.BAYN)[(k0+1):(T)])
dygraph(alpha.BAYN, main = "Rolling Alpha: BAYN") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(beta.BAYN, main = "Rolling Beta: BAYN") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))5.3 BNP
Recursive:
BNP.matrix <- cbind(EX.BNP, EX.MKT)
BNP.matrix <- coredata(BNP.matrix)
BNP.X <- BNP.matrix[,2]
BNP.Y <- BNP.matrix[,1]
k0 <- 99
T <- length(BNP.X)
alpha.BNP <- beta.BNP <- array(dim=(T-k0))
for(i in 1:(T-k0))
{
k <- k0 + i
vX.BNP <- BNP.X[1:k]
vY.BNP <- BNP.Y[1:k]
m2.BNP <- lm(vY.BNP ~ vX.BNP)
alpha.BNP[i] <-m2.BNP$coeff[1]
beta.BNP[i] <- m2.BNP$coeff[2]
}
alpha.BNP <- xts(alpha.BNP, index(EX.BNP)[(k0+1):(T)])
beta.BNP <- xts(beta.BNP, index(EX.BNP)[(k0+1):(T)])
dygraph(alpha.BNP, main = "Recursive Alpha: BNP") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(beta.BNP, main = "Recursive Beta: BNP") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Rolling:
N <- 100
alpha.BNP <- beta.BNP <- array(dim=(T-N+1))
for(i in 1:(T-N+1))
{
vX.BNP <- BNP.X[i:(i+N)]
vY.BNP <- BNP.Y[i:(i+N)]
m2.BNP <- lm(vY.BNP ~ vX.BNP)
alpha.BNP[i] <-m2.BNP$coeff[1]
beta.BNP[i] <- m2.BNP$coeff[2]
}
alpha.BNP <- xts(alpha.BNP, index(EX.BNP)[(k0+1):(T)])
beta.BNP <- xts(beta.BNP, index(EX.BNP)[(k0+1):(T)])
dygraph(alpha.BNP, main = "Rolling Alpha: BNP") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))dygraph(beta.BNP, main = "Rolling Beta: BNP") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))5.4 Índice Sharpe
En el caso de comprobar la constancia o no del índce de Sharpe, vamos a proceder a calcular de forma recursiva el parámetro de forma que se obtendrán distindos índices para cada periodo, comenzando en la observación 100.
De la misma forma se comprobará con Rolling
Recursive Sharpe-ASML:
ASML.sh <- EX.ASML
k0 <- 99
T <- length(ASML.sh)
sharpe.ASML <- array(dim=(T-k0))
for(i in 1:(T-k0))
{
k <- k0 + i
v.ASML.sh <- ASML.sh[1:k]
sharpe.ASML[i] <- SharpeRatio(v.ASML.sh, FUN = "StdDev")
}
sharpe.ASML <- xts(sharpe.ASML, index(EX.ASML)[(k0+1):(T)])
dygraph(sharpe.ASML, main = "Recursive Sharpe Ratio: ASML") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Rolling Sharpe-ASML:
for(i in 1:(T-k0))
{
k <- k0 + i
v.ASML.sh <- ASML.sh[i:k]
sharpe.ASML[i] <- SharpeRatio(v.ASML.sh, FUN = "StdDev")
}
sharpe.ASML <- xts(sharpe.ASML, index(EX.ASML)[(k0+1):(T)])
dygraph(sharpe.ASML, main = "Rolling Sharpe ratio: ASML") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Recursive Sharpe-BAYN:
BAYN.sh <- EX.BAYN
k0 <- 99
T <- length(BAYN.sh)
sharpe.BAYN <- array(dim=(T-k0))
for(i in 1:(T-k0))
{
k <- k0 + i
v.BAYN.sh <- BAYN.sh[1:k]
sharpe.BAYN[i] <- SharpeRatio(v.BAYN.sh, FUN = "StdDev")
}
sharpe.BAYN <- xts(sharpe.BAYN, index(EX.BAYN)[(k0+1):(T)])
dygraph(sharpe.BAYN, main = "Recursive Sharpe ratio: BAYN") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Rolling Sharpe-BAYN:
for(i in 1:(T-k0))
{
k <- k0 + i
v.BAYN.sh <- BAYN.sh[i:k]
sharpe.BAYN[i] <- SharpeRatio(v.BAYN.sh, FUN = "StdDev")
}
sharpe.BAYN <- xts(sharpe.ASML, index(EX.BAYN)[(k0+1):(T)])
dygraph(sharpe.BAYN, main = "Rolling Sharpe ratio: BAYN") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Recursive Sharpe-BNP:
BNP.sh <- EX.BNP
k0 <- 99
T <- length(BNP.sh)
sharpe.BNP <- array(dim=(T-k0))
for(i in 1:(T-k0))
{
k <- k0 + i
v.BNP.sh <- BNP.sh[1:k]
sharpe.BNP[i] <- SharpeRatio(v.BNP.sh, FUN = "StdDev")
}
sharpe.BNP <- xts(sharpe.BNP, index(EX.BNP)[(k0+1):(T)])
dygraph(sharpe.BNP, main = "Recursive Sharpe ratio: BNP") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))Rolling Sharpe-BNP:
for(i in 1:(T-k0))
{
k <- k0 + i
v.BNP.sh <- BNP.sh[i:k]
sharpe.BNP[i] <- SharpeRatio(v.BNP.sh, FUN = "StdDev")
}
sharpe.BNP <- xts(sharpe.BNP, index(EX.BNP)[(k0+1):(T)])
dygraph(sharpe.BNP, main = "Rolling Sharpe ratio: BNP") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))5.5 Índice de Treynor
Dada la construcción del índice sabemos que se utiliza el parámetro beta para su cálculo y, por tanto, podemos aplicar la beta calculada para el procedimiento Rolling y demostrar que no es constante. Por el otro lado la rentabilidad también varía dependiendo del periodo estudado y dado que la beta ni la rentabilidad son constantes, la ratio de Treynor tampoco lo será.
Veamos el enfoque de utilizar la rentabilidad media y la beta de Rolling:
5.5.1 Índice de Treynor-ASML
trey.ASML <-mean(rASML)/beta.ASML
dygraph(trey.ASML, main = "Treynor Ratio con Recursive beta de ASML") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))5.5.2 Índice de Treynor- BAYN
trey.BAYN <-mean(rBAYN)/beta.BAYN
dygraph(trey.BAYN, main = "Treynor Ratio con Recursive beta de BAYN") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))5.5.3 Índice de Treynor- BNP
trey.BNP <-mean(rBNP)/beta.BNP
dygraph(trey.BNP, main = "Treynor Ratio con Recursive beta de BNP") %>% dyRangeSelector(dateWindow = c("2008-01-01", "2017-04-01"))5.6 Alpha de Jensen
Se trata de alpha representado en de los modelos CAPM de los activos por tanto se demuestra también que dicho parámetro no es constante a lo largo del tiempo.
6 Apartado 4: problemas de heteroscedasticidad condicional?
En este apartado analizaremos que si los modelos elegidos en el apartado anterior presentan problemas de heteroscedasticidad condicional.
Hay que señalar que todos los modelos estimados contienen un análisis de los resíduos, veamos los casos para cada activo del modelo seleccionado
6.1 ASML
ggAcf(uhat3.ASML) + labs(title="Residuos de CAPM3")
ggAcf(uhat23.ASML) + labs(title="Residuos al cuadrado de CAPM3")
Box.test(uhat23.ASML, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat23.ASML
## X-squared = 3.3437, df = 12, p-value = 0.9926ArchTest(uhat3.ASML)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat3.ASML
## Chi-squared = 3.1998, df = 12, p-value = 0.994Dados los valores de los contrastes y contemplando los resíduos de este modelo observamos que no hay correlación entre los retardes de los errores y además el test de ARCH no indica esa estructura.
6.2 BAYN
ggAcf(uhat2.BAYN) + labs(title="Residuos CAPM2")
ggAcf(uhat22.BAYN) + labs(title="Residuos al cuadrado de CAPM2")
uhat22.BAYN2 <- uhat22.BAYN*uhat22.BAYN
Box.test(uhat22.BAYN2, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat22.BAYN2
## X-squared = 0.33705, df = 12, p-value = 1ArchTest(uhat2.BAYN)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat2.BAYN
## Chi-squared = 23.286, df = 12, p-value = 0.02539En este caso observamos la no autocorrelación en los errores pero si los efectos ARCH. Dado que tenemos valores atípicos elevados es posible que tras ellos este algún tipo de estructura.
Vamos a ajustar el ARIMA que contangan los resíduos:
fit.uhat2.BAYN<- autoarfima(data =EX.BAYN, ar.max = 2, ma.max = 2,
criterion = "AIC", method = "full")
fit.uhat2.BAYN## $fit
##
## *----------------------------------*
## * ARFIMA Model Fit *
## *----------------------------------*
## Mean Model : ARFIMA(1,0,1)
## Distribution : norm
##
## Optimal Parameters
## ------------------------------------
## Estimate Std. Error t value Pr(>|t|)
## mu 0.042632 0.029731 1.4339 0.151591
## ar1 0.472870 0.211502 2.2358 0.025367
## ma1 -0.542299 0.201938 -2.6855 0.007243
## sigma 1.825515 0.024209 75.4055 0.000000
##
## Robust Standard Errors:
## Estimate Std. Error t value Pr(>|t|)
## mu 0.042632 0.029348 1.4526 0.14633
## ar1 0.472870 0.350357 1.3497 0.17712
## ma1 -0.542299 0.336461 -1.6118 0.10701
## sigma 1.825515 0.097962 18.6349 0.00000
##
## LogLikelihood : -5745.135
##
## Information Criteria
## ------------------------------------
##
## Akaike 4.0444
## Bayes 4.0528
## Shibata 4.0444
## Hannan-Quinn 4.0474
##
## Weighted Ljung-Box Test on Standardized Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 0.02804 0.8670
## Lag[2*(p+q)+(p+q)-1][5] 1.19688 0.9999
## Lag[4*(p+q)+(p+q)-1][9] 5.27827 0.3925
##
## H0 : No serial correlation
##
## Weighted Ljung-Box Test on Standardized Squared Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 18.67 1.557e-05
## Lag[2*(p+q)+(p+q)-1][2] 22.01 1.320e-06
## Lag[4*(p+q)+(p+q)-1][5] 31.22 1.429e-08
##
##
## ARCH LM Tests
## ------------------------------------
## Statistic DoF P-Value
## ARCH Lag[2] 23.66 2 7.280e-06
## ARCH Lag[5] 35.35 5 1.282e-06
## ARCH Lag[10] 43.82 10 3.552e-06
##
## Nyblom stability test
## ------------------------------------
## Joint Statistic: 0.8551
## Individual Statistics:
## mu 0.04045
## ar1 0.04003
## ma1 0.03835
## sigma 0.67980
##
## Asymptotic Critical Values (10% 5% 1%)
## Joint Statistic: 1.07 1.24 1.6
## Individual Statistic: 0.35 0.47 0.75
##
##
## Elapsed time : 0.09501505
##
##
## $rank.matrix
## ar1 ar2 ma1 ma2 im arf AIC converged
## 1 1 0 1 0 1 0 4.044415 1
## 2 1 0 1 0 0 0 4.044430 1
## 3 0 0 1 0 0 0 4.044691 1
## 4 1 0 1 1 1 0 4.044717 1
## 5 1 1 1 0 1 0 4.044735 1
## 6 0 0 1 0 1 0 4.044764 1
## 7 1 1 0 1 1 0 4.044783 1
## 8 0 1 1 1 1 0 4.044792 1
## 9 0 1 1 0 0 0 4.044793 1
## 10 0 0 1 1 0 0 4.044794 1
## 11 1 1 0 1 0 0 4.044822 1
## 12 0 0 1 1 1 0 4.044831 1
## 13 0 1 1 0 1 0 4.044832 1
## 14 0 1 1 1 0 0 4.044835 1
## 15 1 0 0 1 0 0 4.044863 1
## 16 1 1 0 0 0 0 4.044873 1
## 17 1 0 0 1 1 0 4.044905 1
## 18 1 1 0 0 1 0 4.044918 1
## 19 1 1 1 0 0 0 4.044923 1
## 20 1 0 0 0 0 0 4.044952 1
## 21 1 0 1 1 0 0 4.044991 1
## 22 1 0 0 0 1 0 4.045037 1
## 23 1 1 1 1 1 0 4.045486 1
## 24 1 1 1 1 0 0 4.045524 1
## 25 0 1 0 1 0 0 4.048993 1
## 26 0 1 0 1 1 0 4.049145 1
## 27 0 0 0 0 1 0 4.049197 1
## 28 0 1 0 0 0 0 4.049250 1
## 29 0 0 0 1 0 0 4.049262 1
## 30 0 1 0 0 1 0 4.049390 1
## 31 0 0 0 1 1 0 4.049403 1Se trata de un ARIMA(1,0,1)
Ajustemos el GARCH con EX.MKT como variable exógena creando el CAPM:
garch11.BAYN.spec <- ugarchspec(variance.model = list(garchOrder=c(1,1),
model= "apARCH"),
mean.model = list(armaOrder=c(1,1),
external.regressors=EX.MKT),
distribution.model="jsu")
EX.BAYN.garch11.fit <- ugarchfit(spec=garch11.BAYN.spec, data=EX.BAYN,
solver.control=list(trace = 1)) ##
## Iter: 1 fn: 4324.9198 Pars: 0.04890 0.98232 -0.98628 0.92335 0.01272 0.03313 0.96534 0.17555 1.01847 0.11691 1.41445
## Iter: 2 fn: 4324.9198 Pars: 0.04890 0.98232 -0.98628 0.92335 0.01271 0.03313 0.96535 0.17545 1.01826 0.11691 1.41444
## solnp--> Completed in 2 iterationsEX.BAYN.garch11.fit##
## *---------------------------------*
## * GARCH Model Fit *
## *---------------------------------*
##
## Conditional Variance Dynamics
## -----------------------------------
## GARCH Model : apARCH(1,1)
## Mean Model : ARFIMA(1,0,1)
## Distribution : jsu
##
## Optimal Parameters
## ------------------------------------
## Estimate Std. Error t value Pr(>|t|)
## mu 0.048902 0.003271 14.9502 0.000000
## ar1 0.982320 0.003903 251.6852 0.000000
## ma1 -0.986280 0.000045 -21688.3685 0.000000
## mxreg1 0.923350 0.017140 53.8699 0.000000
## omega 0.012710 0.002958 4.2964 0.000017
## alpha1 0.033125 0.003170 10.4509 0.000000
## beta1 0.965346 0.000587 1645.0947 0.000000
## gamma1 0.175454 0.157252 1.1157 0.264530
## delta 1.018261 0.072209 14.1016 0.000000
## skew 0.116909 0.048795 2.3959 0.016579
## shape 1.414444 0.065333 21.6498 0.000000
##
## Robust Standard Errors:
## Estimate Std. Error t value Pr(>|t|)
## mu 0.048902 0.004196 1.1654e+01 0.000000
## ar1 0.982320 0.004113 2.3882e+02 0.000000
## ma1 -0.986280 0.000045 -2.1776e+04 0.000000
## mxreg1 0.923350 0.031205 2.9590e+01 0.000000
## omega 0.012710 0.003526 3.6048e+00 0.000312
## alpha1 0.033125 0.004721 7.0160e+00 0.000000
## beta1 0.965346 0.000773 1.2487e+03 0.000000
## gamma1 0.175454 0.184236 9.5233e-01 0.340929
## delta 1.018261 0.039754 2.5614e+01 0.000000
## skew 0.116909 0.050748 2.3037e+00 0.021240
## shape 1.414444 0.086428 1.6366e+01 0.000000
##
## LogLikelihood : -4324.92
##
## Information Criteria
## ------------------------------------
##
## Akaike 3.0502
## Bayes 3.0733
## Shibata 3.0502
## Hannan-Quinn 3.0585
##
## Weighted Ljung-Box Test on Standardized Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 2.968 0.08494
## Lag[2*(p+q)+(p+q)-1][5] 3.903 0.08468
## Lag[4*(p+q)+(p+q)-1][9] 5.278 0.39257
## d.o.f=2
## H0 : No serial correlation
##
## Weighted Ljung-Box Test on Standardized Squared Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 18.17 2.020e-05
## Lag[2*(p+q)+(p+q)-1][5] 18.41 5.281e-05
## Lag[4*(p+q)+(p+q)-1][9] 18.68 4.364e-04
## d.o.f=2
##
## Weighted ARCH LM Tests
## ------------------------------------
## Statistic Shape Scale P-Value
## ARCH Lag[3] 0.3254 0.500 2.000 0.5684
## ARCH Lag[5] 0.3742 1.440 1.667 0.9200
## ARCH Lag[7] 0.4798 2.315 1.543 0.9802
##
## Nyblom stability test
## ------------------------------------
## Joint Statistic: 5.3931
## Individual Statistics:
## mu 0.25792
## ar1 0.10757
## ma1 0.10634
## mxreg1 3.40026
## omega 0.62073
## alpha1 0.63672
## beta1 0.61846
## gamma1 0.03871
## delta 0.23885
## skew 0.22323
## shape 0.18656
##
## Asymptotic Critical Values (10% 5% 1%)
## Joint Statistic: 2.49 2.75 3.27
## Individual Statistic: 0.35 0.47 0.75
##
## Sign Bias Test
## ------------------------------------
## t-value prob sig
## Sign Bias 0.3800 0.7039559
## Negative Sign Bias 0.7169 0.4734927
## Positive Sign Bias 3.3129 0.0009348 ***
## Joint Effect 12.2569 0.0065531 ***
##
##
## Adjusted Pearson Goodness-of-Fit Test:
## ------------------------------------
## group statistic p-value(g-1)
## 1 20 20.01 0.3939
## 2 30 23.61 0.7480
## 3 40 34.99 0.6534
## 4 50 49.42 0.4563
##
##
## Elapsed time : 32.42306Veamos la comparación entre los parámetros obtenidos en el modelo CAPM y el GARCH(1,1):
alpha.BAYN.GARCH <-as.data.frame(coef(EX.BAYN.garch11.fit)[6])
beta.BAYN.GARCH <- as.data.frame(coef(EX.BAYN.garch11.fit)[7])
garch.parmeters.BAYN <- cbind(alpha.BAYN.GARCH,beta.BAYN.GARCH, alpha2.BAYN,beta2.BAYN)
colnames(garch.parmeters.BAYN)<- c("Alpha-GARCH", "Beta-GARCH", "Alpha-CAPM2", "Beta-CAPM2")
rownames(garch.parmeters.BAYN)<- c("Parámetros")
garch.parmeters.BAYN## Alpha-GARCH Beta-GARCH Alpha-CAPM2 Beta-CAPM2
## Parámetros 0.03312527 0.965346 0.05737158 0.8602435Como podemos observar no hay campbio sistancial
Veamos los errores del GARCH y su sigma.
uhat.garch.BAYN <- residuals(EX.BAYN.garch11.fit)
uhat.garch.BAYN2 <- uhat.garch.BAYN*uhat.garch.BAYN
ArchTest(uhat.garch.BAYN)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat.garch.BAYN
## Chi-squared = 24.424, df = 12, p-value = 0.0178Box.test(uhat.garch.BAYN2,lag=12, type = "Ljung-Box")##
## Box-Ljung test
##
## data: uhat.garch.BAYN2
## X-squared = 25.015, df = 12, p-value = 0.01475plot.ts(residuals(EX.BAYN.garch11.fit), ylab="e(t)", col="blue")
abline(h=0)
plot.ts(sigma(EX.BAYN.garch11.fit), ylab="sigma(t)", col="blue")
plot(EX.BAYN.garch11.fit, which=9)
plot(EX.BAYN.garch11.fit, which=10)
plot(EX.BAYN.garch11.fit, which=11)
plot(EX.BAYN.garch11.fit, which=12)
La ACF parece ser una normal por lo que nosquedamos con ese modelo, aunque la ACF de los rsiduos presenta elevados valores atípicos. El primer LAG significativo no desaparece al ajustar los distintos ARIMA.
6.2.1 índices BAYN
sharpe.garch.BAYN <- mean(EX.BAYN)/mean(sigma(EX.BAYN.garch11.fit))
treynor.garch.BAYN <- mean(EX.BAYN)/as.data.frame(coef(EX.BAYN.garch11.fit)[6])
jensen.garch.BAYN <- as.data.frame(coef(EX.BAYN.garch11.fit)[5])
ind.capm.BAYN<- cbind(sharpe.ratio.BAYN,treynor.ratio.BAYN,jensen.ratio.BAYN)
colnames(ind.capm.BAYN) <- c("Sharpe", "Treynor", "Jensen")
ind.garch.BAYN<- cbind(sharpe.garch.BAYN,treynor.garch.BAYN,jensen.garch.BAYN)
colnames(ind.garch.BAYN) <- c("Sharpe", "Treynor", "Jensen")
compare.index.BAYN <- rbind(ind.capm.BAYN,ind.garch.BAYN)
rownames(compare.index.BAYN)<- c("Ratios CAPM", "Ratios GARCH")
compare.index.BAYN## Sharpe Treynor Jensen
## Ratios CAPM 0.02318896 0.04937026 0.05737158
## Ratios GARCH 0.03535170 1.28211641 0.01271041Se observa diferencias significativas entre las distintas ratios.
6.3 BNP
ggAcf(uhat2.BNP) + labs(title="Residuos CAPM2")
ggAcf(uhat22.BNP) + labs(title="Residuos al cuadrado de CAPM2")
uhat22.BNP2 <- uhat22.BNP*uhat22.BNP
Box.test(uhat22.BNP2, lag=12, type="Ljung-Box")##
## Box-Ljung test
##
## data: uhat22.BNP2
## X-squared = 35.358, df = 12, p-value = 0.0004105ArchTest(uhat2.BNP)##
## ARCH LM-test; Null hypothesis: no ARCH effects
##
## data: uhat2.BNP
## Chi-squared = 386.48, df = 12, p-value < 2.2e-16En este caso observamos la autocorrelación en los errores como los efectos ARCH.
Vamos a ajustar el ARIMA que contangan los resíduos:
fit.uhat2.BNP<- autoarfima(data =EX.BNP, ar.max = 3, ma.max = 3,
criterion = "AIC", method = "full")
fit.uhat2.BNP## $fit
##
## *----------------------------------*
## * ARFIMA Model Fit *
## *----------------------------------*
## Mean Model : ARFIMA(1,0,3)
## Distribution : norm
##
## Optimal Parameters
## ------------------------------------
## Estimate Std. Error t value Pr(>|t|)
## ar1 0.532643 0.127180 4.1881 2.8e-05
## ma1 -0.522312 0.127628 -4.0925 4.3e-05
## ma2 0.000000 NA NA NA
## ma3 -0.098662 0.019702 -5.0077 1.0e-06
## sigma 2.544527 0.033769 75.3515 0.0e+00
##
## Robust Standard Errors:
## Estimate Std. Error t value Pr(>|t|)
## ar1 0.532643 0.224869 2.3687 0.017852
## ma1 -0.522312 0.231260 -2.2585 0.023911
## ma2 0.000000 NA NA NA
## ma3 -0.098662 0.039977 -2.4679 0.013590
## sigma 2.544527 0.077320 32.9089 0.000000
##
## LogLikelihood : -6687.889
##
## Information Criteria
## ------------------------------------
##
## Akaike 4.7076
## Bayes 4.7160
## Shibata 4.7076
## Hannan-Quinn 4.7106
##
## Weighted Ljung-Box Test on Standardized Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 0.008781 0.9253
## Lag[2*(p+q)+(p+q)-1][11] 3.457545 1.0000
## Lag[4*(p+q)+(p+q)-1][19] 10.233338 0.4299
##
## H0 : No serial correlation
##
## Weighted Ljung-Box Test on Standardized Squared Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 76.63 0
## Lag[2*(p+q)+(p+q)-1][2] 156.14 0
## Lag[4*(p+q)+(p+q)-1][5] 346.90 0
##
##
## ARCH LM Tests
## ------------------------------------
## Statistic DoF P-Value
## ARCH Lag[2] 204.5 2 0
## ARCH Lag[5] 319.3 5 0
## ARCH Lag[10] 356.5 10 0
##
## Nyblom stability test
## ------------------------------------
## Joint Statistic: 2.8013
## Individual Statistics:
## ar1 0.02600
## ma1 0.01911
## ma3 0.02290
## sigma 2.59563
##
## Asymptotic Critical Values (10% 5% 1%)
## Joint Statistic: 1.07 1.24 1.6
## Individual Statistic: 0.35 0.47 0.75
##
##
## Elapsed time : 0.06283402
##
##
## $rank.matrix
## ar1 ar2 ar3 ma1 ma2 ma3 im arf AIC converged
## 1 1 0 0 1 0 1 0 0 4.707625 1
## 2 0 1 0 0 1 1 0 0 4.707753 1
## 3 1 0 1 1 0 1 0 0 4.707929 1
## 4 0 1 1 0 1 0 0 0 4.708082 1
## 5 1 0 1 1 0 0 0 0 4.708161 1
## 6 1 1 1 1 1 0 0 0 4.708166 1
## 7 0 0 1 0 0 1 0 0 4.708187 1
## 8 1 1 0 1 1 1 0 0 4.708197 1
## 9 1 1 0 1 0 1 0 0 4.708317 1
## 10 1 0 0 1 0 1 1 0 4.708319 1
## 11 1 0 0 1 1 1 0 0 4.708328 1
## 12 0 0 0 0 0 1 0 0 4.708367 1
## 13 1 1 0 0 1 1 0 0 4.708376 1
## 14 0 1 0 1 1 1 0 0 4.708425 1
## 15 0 1 1 0 1 1 0 0 4.708429 1
## 16 0 1 0 0 1 1 1 0 4.708447 1
## 17 1 0 1 1 0 1 1 0 4.708623 1
## 18 1 0 1 1 1 1 0 0 4.708632 1
## 19 1 1 1 1 0 1 0 0 4.708632 1
## 20 1 1 1 0 1 0 0 0 4.708641 1
## 21 0 1 1 1 1 0 0 0 4.708735 1
## 22 0 1 1 0 1 0 1 0 4.708777 1
## 23 1 1 1 1 0 0 0 0 4.708855 1
## 24 1 0 1 1 0 0 1 0 4.708856 1
## 25 1 1 1 1 1 0 1 0 4.708861 1
## 26 1 0 1 1 1 0 0 0 4.708864 1
## 27 1 1 1 1 1 1 0 0 4.708869 1
## 28 0 0 1 0 1 1 0 0 4.708874 1
## 29 0 0 1 0 0 1 1 0 4.708882 1
## 30 1 0 1 0 0 1 0 0 4.708884 1
## 31 0 0 1 1 0 1 0 0 4.708888 1
## 32 0 1 1 0 0 1 0 0 4.708890 1
## 33 1 1 0 1 1 1 1 0 4.708892 1
## 34 1 0 0 0 0 1 0 0 4.708940 1
## 35 0 1 0 0 0 1 0 0 4.708989 1
## 36 0 0 0 1 0 1 0 0 4.708998 1
## 37 1 1 0 1 0 1 1 0 4.709011 1
## 38 1 0 0 1 1 1 1 0 4.709024 1
## 39 0 0 0 0 1 1 0 0 4.709040 1
## 40 0 0 0 0 0 1 1 0 4.709063 1
## 41 1 1 0 0 1 1 1 0 4.709071 1
## 42 1 1 1 0 1 1 0 0 4.709077 1
## 43 0 0 1 0 0 0 0 0 4.709098 1
## 44 0 1 1 1 1 1 0 0 4.709119 1
## 45 0 1 0 1 1 1 1 0 4.709119 1
## 46 0 1 1 0 1 1 1 0 4.709123 1
## 47 1 0 1 1 1 1 1 0 4.709326 1
## 48 1 1 1 1 0 1 1 0 4.709326 1
## 49 1 1 1 0 1 0 1 0 4.709336 1
## 50 0 1 1 1 1 0 1 0 4.709430 1
## 51 1 1 1 1 0 0 1 0 4.709549 1
## 52 1 0 1 1 1 0 1 0 4.709558 1
## 53 1 1 1 1 1 1 1 0 4.709563 1
## 54 1 1 0 0 0 1 0 0 4.709564 1
## 55 0 0 1 0 1 1 1 0 4.709569 1
## 56 1 0 1 0 1 1 0 0 4.709572 1
## 57 0 0 1 1 1 1 0 0 4.709573 1
## 58 1 0 1 0 0 1 1 0 4.709579 1
## 59 0 0 1 1 0 1 1 0 4.709583 1
## 60 0 1 1 0 0 1 1 0 4.709586 1
## 61 1 1 1 0 0 1 0 0 4.709587 1
## 62 0 1 1 1 0 1 0 0 4.709591 1
## 63 1 0 1 0 0 0 0 0 4.709603 1
## 64 1 0 0 0 1 1 0 0 4.709612 1
## 65 0 1 0 1 0 1 0 0 4.709618 1
## 66 1 0 0 0 0 1 1 0 4.709636 1
## 67 0 1 1 0 0 0 0 0 4.709643 1
## 68 0 0 1 1 0 0 0 0 4.709653 1
## 69 0 0 0 1 1 1 0 0 4.709668 1
## 70 0 1 0 0 0 1 1 0 4.709685 1
## 71 0 0 0 1 0 1 1 0 4.709693 1
## 72 0 0 1 0 1 0 0 0 4.709710 1
## 73 0 0 0 0 1 1 1 0 4.709735 1
## 74 1 1 1 0 1 1 1 0 4.709771 1
## 75 0 0 1 0 0 0 1 0 4.709794 1
## 76 0 1 1 1 1 1 1 0 4.709813 1
## 77 1 1 1 0 0 0 0 0 4.710154 1
## 78 0 1 1 1 0 0 0 0 4.710199 1
## 79 1 0 1 0 1 0 0 0 4.710219 1
## 80 1 1 0 0 0 1 1 0 4.710260 1
## 81 0 0 1 1 1 0 0 0 4.710264 1
## 82 1 0 1 0 1 1 1 0 4.710268 1
## 83 0 0 1 1 1 1 1 0 4.710269 1
## 84 1 1 1 0 0 1 1 0 4.710283 1
## 85 0 1 1 1 0 1 1 0 4.710286 1
## 86 1 0 1 0 0 0 1 0 4.710300 1
## 87 1 0 0 0 1 1 1 0 4.710309 1
## 88 0 1 0 1 0 1 1 0 4.710314 1
## 89 0 1 1 0 0 0 1 0 4.710340 1
## 90 0 0 1 1 0 0 1 0 4.710349 1
## 91 0 0 0 1 1 1 1 0 4.710363 1
## 92 0 0 1 0 1 0 1 0 4.710407 1
## 93 1 1 1 0 0 0 1 0 4.710850 1
## 94 0 1 1 1 0 0 1 0 4.710896 1
## 95 1 0 1 0 1 0 1 0 4.710915 1
## 96 0 0 1 1 1 0 1 0 4.710960 1
## 97 1 1 0 1 1 0 0 0 4.715757 1
## 98 1 0 0 1 1 0 0 0 4.716227 1
## 99 1 1 0 1 1 0 1 0 4.716455 1
## 100 1 1 0 1 0 0 0 0 4.716496 1
## 101 0 1 0 1 1 0 0 0 4.716772 1
## 102 1 0 0 1 1 0 1 0 4.716925 1
## 103 1 1 0 0 1 0 0 0 4.716945 1
## 104 1 1 0 1 0 0 1 0 4.717193 1
## 105 0 1 0 1 1 0 1 0 4.717469 1
## 106 1 1 0 0 1 0 1 0 4.717643 1
## 107 1 0 0 0 0 0 0 0 4.718466 1
## 108 0 0 0 1 0 0 0 0 4.718470 1
## 109 0 0 0 0 1 0 0 0 4.718498 1
## 110 0 1 0 0 0 0 0 0 4.718504 1
## 111 0 0 0 0 0 0 1 0 4.718625 1
## 112 0 0 0 1 1 0 0 0 4.718970 1
## 113 1 0 0 0 1 0 0 0 4.719009 1
## 114 0 1 0 1 0 0 0 0 4.719015 1
## 115 1 1 0 0 0 0 0 0 4.719045 1
## 116 0 1 0 0 1 0 0 0 4.719140 1
## 117 1 0 0 0 0 0 1 0 4.719164 1
## 118 0 0 0 1 0 0 1 0 4.719168 1
## 119 1 0 0 1 0 0 0 0 4.719171 1
## 120 0 0 0 0 1 0 1 0 4.719196 1
## 121 0 1 0 0 0 0 1 0 4.719201 1
## 122 0 0 0 1 1 0 1 0 4.719667 1
## 123 1 0 0 0 1 0 1 0 4.719707 1
## 124 0 1 0 1 0 0 1 0 4.719712 1
## 125 1 1 0 0 0 0 1 0 4.719743 1
## 126 0 1 0 0 1 0 1 0 4.719838 1
## 127 1 0 0 1 0 0 1 0 4.719868 1Se trata de un ARIMA(1,0,1)
Ajustemos el GARCH con EX.MKT como variable exógena creando el CAPM:
Nota: aunque el ajuste autoarfima marca un ARIMA(1,0,3), los parámetros no son significativos, tras reajustes se observa que el parámetro AR(1) es significativo.
garch11.BNP.spec <- ugarchspec(variance.model = list(garchOrder=c(1,1)),
mean.model = list(armaOrder=c(1,0),
external.regressors=EX.MKT),
distribution.model="sstd")
EX.BNP.garch11.fit <- ugarchfit(spec=garch11.BNP.spec, data=EX.BNP,
solver.control=list(trace = 1)) ##
## Iter: 1 fn: 4359.2513 Pars: 0.02875 0.05359 1.31046 0.01409 0.07319 0.92374 1.08822 4.75504
## Iter: 2 fn: 4359.2513 Pars: 0.02872 0.05359 1.31046 0.01408 0.07320 0.92375 1.08817 4.75471
## solnp--> Completed in 2 iterationsEX.BNP.garch11.fit##
## *---------------------------------*
## * GARCH Model Fit *
## *---------------------------------*
##
## Conditional Variance Dynamics
## -----------------------------------
## GARCH Model : sGARCH(1,1)
## Mean Model : ARFIMA(1,0,0)
## Distribution : sstd
##
## Optimal Parameters
## ------------------------------------
## Estimate Std. Error t value Pr(>|t|)
## mu 0.028720 0.019539 1.4699 0.141585
## ar1 0.053588 0.018446 2.9051 0.003671
## mxreg1 1.310459 0.016890 77.5868 0.000000
## omega 0.014076 0.005341 2.6352 0.008408
## alpha1 0.073198 0.013062 5.6040 0.000000
## beta1 0.923747 0.013074 70.6543 0.000000
## skew 1.088174 0.028134 38.6781 0.000000
## shape 4.754711 0.436213 10.9000 0.000000
##
## Robust Standard Errors:
## Estimate Std. Error t value Pr(>|t|)
## mu 0.028720 0.019686 1.4589 0.144597
## ar1 0.053588 0.018170 2.9492 0.003186
## mxreg1 1.310459 0.029471 44.4661 0.000000
## omega 0.014076 0.006451 2.1818 0.029121
## alpha1 0.073198 0.019440 3.7654 0.000166
## beta1 0.923747 0.019089 48.3905 0.000000
## skew 1.088174 0.030460 35.7253 0.000000
## shape 4.754711 0.467590 10.1686 0.000000
##
## LogLikelihood : -4359.251
##
## Information Criteria
## ------------------------------------
##
## Akaike 3.0723
## Bayes 3.0890
## Shibata 3.0723
## Hannan-Quinn 3.0783
##
## Weighted Ljung-Box Test on Standardized Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 0.8355 0.3607
## Lag[2*(p+q)+(p+q)-1][2] 0.8366 0.8326
## Lag[4*(p+q)+(p+q)-1][5] 1.7988 0.7683
## d.o.f=1
## H0 : No serial correlation
##
## Weighted Ljung-Box Test on Standardized Squared Residuals
## ------------------------------------
## statistic p-value
## Lag[1] 0.4998 0.4796
## Lag[2*(p+q)+(p+q)-1][5] 1.5763 0.7211
## Lag[4*(p+q)+(p+q)-1][9] 2.6114 0.8210
## d.o.f=2
##
## Weighted ARCH LM Tests
## ------------------------------------
## Statistic Shape Scale P-Value
## ARCH Lag[3] 0.1723 0.500 2.000 0.6781
## ARCH Lag[5] 1.4603 1.440 1.667 0.6028
## ARCH Lag[7] 1.8567 2.315 1.543 0.7476
##
## Nyblom stability test
## ------------------------------------
## Joint Statistic: 4.5771
## Individual Statistics:
## mu 0.08178
## ar1 0.21932
## mxreg1 1.64957
## omega 1.06128
## alpha1 0.52974
## beta1 0.76913
## skew 0.41568
## shape 1.40195
##
## Asymptotic Critical Values (10% 5% 1%)
## Joint Statistic: 1.89 2.11 2.59
## Individual Statistic: 0.35 0.47 0.75
##
## Sign Bias Test
## ------------------------------------
## t-value prob sig
## Sign Bias 0.79491 0.42673
## Negative Sign Bias 1.77705 0.07567 *
## Positive Sign Bias 0.05616 0.95522
## Joint Effect 3.16490 0.36689
##
##
## Adjusted Pearson Goodness-of-Fit Test:
## ------------------------------------
## group statistic p-value(g-1)
## 1 20 4.858 0.9995
## 2 30 15.210 0.9833
## 3 40 32.343 0.7657
## 4 50 38.024 0.8720
##
##
## Elapsed time : 14.04797Veamos la comparación entre los parámetros obtenidos en el modelo CAPM y el GARCH(1,1):
alpha.BNP.GARCH <-as.data.frame(coef(EX.BNP.garch11.fit)[5])
beta.BNP.GARCH <- as.data.frame(coef(EX.BNP.garch11.fit)[6])
garch.parmeters.BNP <- cbind(alpha.BNP.GARCH,beta.BNP.GARCH, alpha2.BNP,beta2.BNP)
colnames(garch.parmeters.BNP)<- c("Alpha-GARCH", "Beta-GARCH", "Alpha-CAPM2", "Beta-CAPM2")
rownames(garch.parmeters.BNP)<- c("Parámetros")
garch.parmeters.BNP## Alpha-GARCH Beta-GARCH Alpha-CAPM2 Beta-CAPM2
## Parámetros 0.07319824 0.9237467 0.03096202 1.421348plot(EX.BNP.garch11.fit, which=9)
plot(EX.BNP.garch11.fit, which=10)
plot(EX.BNP.garch11.fit, which=11)
plot(EX.BNP.garch11.fit, which=12)
En este caso observamos que la ACF parece ser ruido blanco.
6.3.1 índices BNP
sharpe.garch.BNP <- mean(EX.BNP)/mean(sigma(EX.BNP.garch11.fit))
treynor.garch.BNP <- mean(EX.BNP)/as.data.frame(coef(EX.BNP.garch11.fit)[6])
jensen.garch.BNP <- as.data.frame(coef(EX.BNP.garch11.fit)[5])
ind.capm.BNP<- cbind(sharpe.ratio.BNP,treynor.ratio.BNP,jensen.ratio.BNP)
colnames(ind.capm.BNP) <- c("Sharpe", "Treynor", "Jensen")
ind.garch.BNP<- cbind(sharpe.garch.BNP,treynor.garch.BNP,jensen.garch.BNP)
colnames(ind.garch.BNP) <- c("Sharpe", "Treynor", "Jensen")
compare.index.BNP <- rbind(ind.capm.BNP,ind.garch.BNP)
rownames(compare.index.BNP)<- c("Ratios CAPM", "Ratios GARCH")
compare.index.BNP## Sharpe Treynor Jensen
## Ratios CAPM 0.002477528 0.004461577 0.03096202
## Ratios GARCH 0.004708877 0.006864925 0.07319824Como se puede observar, las ratio apriximadamente se duplican.
7 Apartado 5: relación de cointegración?
En este apartado comprobaremos si existe alguna relación de cointegración entre los parámetros.
Vamos a calcular los logaritmos de los precios:
ldata <- log(fdata)
ldata <- na.omit(ldata)
lASML <- ldata$ASML
lBAYN <- ldata$BAYN
lBNP <- ldata$BNPVamos a convertir los datos a zoo:
asml<-as.zoo(lASML)
bayn<-as.zoo(lBAYN)
bnp<-as.zoo(lBNP)7.1 Engle-Granger
7.1.1 ASML-BAYN~1
Primer paso:
e1 <- dynlm(asml ~ bayn)
print(summary(e1))##
## Time series regression with "zoo" data:
## Start = 2005-01-03, End = 2017-04-07
##
## Call:
## dynlm(formula = asml ~ bayn)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.61288 -0.13936 -0.01707 0.14359 0.54364
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.718403 0.030570 -56.21 <2e-16 ***
## bayn 1.319165 0.007674 171.91 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2198 on 3038 degrees of freedom
## Multiple R-squared: 0.9068, Adjusted R-squared: 0.9068
## F-statistic: 2.955e+04 on 1 and 3038 DF, p-value: < 2.2e-16tsdisplay(e1$residuals)
z1 <- e1$residualsLa ACF sugiere que los residuos no son estacionarios. Puede haber una raíz unitaria.
Segundo paso:
Aplicamos la ADF a \(z1\)
- Con constante y tendencia
z1.df3 <- ur.df(z1, type="trend", selectlags = "BIC")
print(summary(z1.df3))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression trend
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.32684 -0.01357 -0.00058 0.01300 0.19345
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.102e-03 9.597e-04 -1.149 0.25080
## z.lag.1 -7.026e-03 2.210e-03 -3.179 0.00149 **
## tt 7.906e-07 5.525e-07 1.431 0.15252
## z.diff.lag -3.619e-02 1.815e-02 -1.994 0.04621 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02557 on 3034 degrees of freedom
## Multiple R-squared: 0.004982, Adjusted R-squared: 0.003998
## F-statistic: 5.064 on 3 and 3034 DF, p-value: 0.001688
##
##
## Value of test-statistic is: -3.1792 3.4811 5.1979
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau3 -3.96 -3.41 -3.12
## phi2 6.09 4.68 4.03
## phi3 8.27 6.25 5.34plot(z1.df3)
NO podemos rechazar la \(H_0\) por lo que parace que existe la raíz unitaria.
Dado el valor crítico de \(\phi_3\) podemos eliminar la tendencia.
z1.df2 <- ur.df(z1, type="drift", selectlags = "BIC")
print(summary(z1.df2))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression drift
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.32743 -0.01326 -0.00052 0.01291 0.19332
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 9.991e-05 4.640e-04 0.215 0.82953
## z.lag.1 -6.114e-03 2.116e-03 -2.889 0.00389 **
## z.diff.lag -3.642e-02 1.815e-02 -2.007 0.04484 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02557 on 3035 degrees of freedom
## Multiple R-squared: 0.004311, Adjusted R-squared: 0.003655
## F-statistic: 6.57 on 2 and 3035 DF, p-value: 0.001422
##
##
## Value of test-statistic is: -2.8888 4.1962
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau2 -3.43 -2.86 -2.57
## phi1 6.43 4.59 3.78plot(z1.df2)
Parece que no hay raíz unitaria por lo que existe la cointegración
La constante parece no ser signiicativa, hay que tener en cuanta que a falta de constante los test se vuelven inestables.
Veamos el modelo sin constante:
z1.df1 <- ur.df(z1, type="none", selectlags = "BIC")
print(summary(z1.df1))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression none
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.32733 -0.01316 -0.00042 0.01301 0.19342
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## z.lag.1 -0.006114 0.002116 -2.889 0.00389 **
## z.diff.lag -0.036407 0.018145 -2.006 0.04489 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02557 on 3036 degrees of freedom
## Multiple R-squared: 0.00431, Adjusted R-squared: 0.003654
## F-statistic: 6.571 on 2 and 3036 DF, p-value: 0.001421
##
##
## Value of test-statistic is: -2.8894
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau1 -2.58 -1.95 -1.62plot(z1.df1) Parece que no hay raíz unitaria por lo que existe la cointegración
7.1.2 ASML-BNP~2
Primer paso:
e2 <- dynlm(asml ~ bnp)
print(summary(e2))##
## Time series regression with "zoo" data:
## Start = 2005-01-03, End = 2017-04-07
##
## Call:
## dynlm(formula = asml ~ bnp)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.17165 -0.68224 -0.01717 0.65174 1.12240
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.40904 0.19035 7.402 1.72e-13 ***
## bnp 0.56535 0.05155 10.968 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.7059 on 3038 degrees of freedom
## Multiple R-squared: 0.03809, Adjusted R-squared: 0.03777
## F-statistic: 120.3 on 1 and 3038 DF, p-value: < 2.2e-16tsdisplay(e2$residuals)
z2 <- e2$residualsLa ACF sugiere que los residuos no son estacionarios. Puede haber una raíz unitaria.
Segundo paso:
Aplicamos la ADF a \(z2\)
- Con constante y tendencia
z2.df3 <- ur.df(z2, type="trend", selectlags = "BIC")
print(summary(z2.df3))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression trend
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.221551 -0.010330 -0.000326 0.010472 0.199526
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -9.884e-03 2.993e-03 -3.303 0.000968 ***
## z.lag.1 -8.908e-03 2.428e-03 -3.669 0.000248 ***
## tt 6.983e-06 1.953e-06 3.576 0.000354 ***
## z.diff.lag -1.612e-03 1.815e-02 -0.089 0.929222
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02035 on 3034 degrees of freedom
## Multiple R-squared: 0.004454, Adjusted R-squared: 0.003469
## F-statistic: 4.524 on 3 and 3034 DF, p-value: 0.003595
##
##
## Value of test-statistic is: -3.6687 5.8123 6.7302
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau3 -3.96 -3.41 -3.12
## phi2 6.09 4.68 4.03
## phi3 8.27 6.25 5.34plot(z2.df3)
NO podemos rechazar la \(H_0\) por lo que parace que existe la raíz unitaria.
Dado el valor crítico de \(\phi_3\) podemos eliminar la tendencia.
z2.df2 <- ur.df(z2, type="drift", selectlags = "BIC")
print(summary(z2.df2))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression drift
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.22308 -0.01032 -0.00045 0.01046 0.20048
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.0007366 0.0003702 1.990 0.0467 *
## z.lag.1 -0.0004287 0.0005245 -0.817 0.4138
## z.diff.lag -0.0058534 0.0181478 -0.323 0.7471
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02039 on 3035 degrees of freedom
## Multiple R-squared: 0.0002569, Adjusted R-squared: -0.0004019
## F-statistic: 0.3899 on 2 and 3035 DF, p-value: 0.6771
##
##
## Value of test-statistic is: -0.8173 2.3146
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau2 -3.43 -2.86 -2.57
## phi1 6.43 4.59 3.78plot(z2.df2)
Parece que hay raíz unitaria por lo que no existe la cointegración
La constante parece no ser signiicativa, hay que tener en cuanta que a falta de constante los test se vuelven inestables.
Veamos el modelo sin constante:
z2.df1 <- ur.df(z2, type="none", selectlags = "BIC")
print(summary(z2.df1))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression none
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.222376 -0.009612 0.000294 0.011177 0.201192
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## z.lag.1 -0.0004293 0.0005248 -0.818 0.413
## z.diff.lag -0.0045739 0.0181452 -0.252 0.801
##
## Residual standard error: 0.0204 on 3036 degrees of freedom
## Multiple R-squared: 0.0002433, Adjusted R-squared: -0.0004153
## F-statistic: 0.3694 on 2 and 3036 DF, p-value: 0.6912
##
##
## Value of test-statistic is: -0.818
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau1 -2.58 -1.95 -1.62plot(z2.df1)
Parece que hay raíz unitaria por lo que no existe la cointegración
7.1.3 BAYN-BNP~3
Primer paso:
e3 <- dynlm(bayn ~ bnp)
print(summary(e3))##
## Time series regression with "zoo" data:
## Start = 2005-01-03, End = 2017-04-07
##
## Call:
## dynlm(formula = bayn ~ bnp)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.86694 -0.34484 -0.03923 0.44166 0.87529
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.29535 0.13152 9.849 <2e-16 ***
## bnp 0.72045 0.03562 20.229 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4877 on 3038 degrees of freedom
## Multiple R-squared: 0.1187, Adjusted R-squared: 0.1184
## F-statistic: 409.2 on 1 and 3038 DF, p-value: < 2.2e-16tsdisplay(e3$residuals)
z3 <- e3$residualsLa ACF sugiere que los residuos no son estacionarios. Puede haber una raíz unitaria.
Segundo paso:
Aplicamos la ADF a \(z3\)
- Con constante y tendencia
z3.df3 <- ur.df(z3, type="trend", selectlags = "BIC")
print(summary(z3.df3))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression trend
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.144507 -0.009220 -0.000100 0.009163 0.262314
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.637e-03 2.076e-03 -2.715 0.00667 **
## z.lag.1 -7.925e-03 2.422e-03 -3.272 0.00108 **
## tt 3.981e-06 1.346e-06 2.957 0.00314 **
## z.diff.lag 5.366e-03 1.817e-02 0.295 0.76778
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.01898 on 3034 degrees of freedom
## Multiple R-squared: 0.003633, Adjusted R-squared: 0.002647
## F-statistic: 3.687 on 3 and 3034 DF, p-value: 0.0115
##
##
## Value of test-statistic is: -3.2719 4.1738 5.5289
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau3 -3.96 -3.41 -3.12
## phi2 6.09 4.68 4.03
## phi3 8.27 6.25 5.34plot(z3.df3)
NO podemos rechazar la \(H_0\) por lo que parace que existe la raíz unitaria.
Dado el valor crítico de \(\phi_3\) podemos eliminar la tendencia.
z3.df2 <- ur.df(z3, type="drift", selectlags = "BIC")
print(summary(z3.df2))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression drift
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.142952 -0.009028 -0.000065 0.009163 0.264502
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.0004166 0.0003448 1.208 0.227
## z.lag.1 -0.0010750 0.0007072 -1.520 0.129
## z.diff.lag 0.0013972 0.0181431 0.077 0.939
##
## Residual standard error: 0.019 on 3035 degrees of freedom
## Multiple R-squared: 0.000762, Adjusted R-squared: 0.0001035
## F-statistic: 1.157 on 2 and 3035 DF, p-value: 0.3145
##
##
## Value of test-statistic is: -1.5201 1.8854
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau2 -3.43 -2.86 -2.57
## phi1 6.43 4.59 3.78plot(z3.df2)
Parece que hay raíz unitaria por lo que no existe la cointegración
La constante parece no ser signiicativa, hay que tener en cuanta que a falta de constante los test se vuelven inestables.
Veamos el modelo sin constante:
z3.df1 <- ur.df(z3, type="none", selectlags = "BIC")
print(summary(z3.df1))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression none
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.142605 -0.008610 0.000353 0.009583 0.264918
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## z.lag.1 -0.0010751 0.0007072 -1.520 0.129
## z.diff.lag 0.0018716 0.0181402 0.103 0.918
##
## Residual standard error: 0.019 on 3036 degrees of freedom
## Multiple R-squared: 0.0007629, Adjusted R-squared: 0.0001046
## F-statistic: 1.159 on 2 and 3036 DF, p-value: 0.314
##
##
## Value of test-statistic is: -1.5201
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau1 -2.58 -1.95 -1.62plot(z3.df1)
Parece que hay raíz unitaria por lo que no existe la cointegración
7.1.4 BAYN -ASML~4
Primer paso:
e4 <- dynlm(bayn ~ asml)
print(summary(e4))##
## Time series regression with "zoo" data:
## Start = 2005-01-03, End = 2017-04-07
##
## Call:
## dynlm(formula = bayn ~ asml)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.4147 -0.1199 0.0015 0.1290 0.4213
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.549408 0.014257 108.7 <2e-16 ***
## asml 0.687391 0.003999 171.9 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.1586 on 3038 degrees of freedom
## Multiple R-squared: 0.9068, Adjusted R-squared: 0.9068
## F-statistic: 2.955e+04 on 1 and 3038 DF, p-value: < 2.2e-16tsdisplay(e4$residuals)
z4 <- e4$residualsLa ACF sugiere que los residuos no son estacionarios. Puede haber una raíz unitaria.
Segundo paso:
Aplicamos la ADF a \(z4\)
- Con constante y tendencia
z4.df3 <- ur.df(z4, type="trend", selectlags = "BIC")
print(summary(z4.df3))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression trend
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.136194 -0.009377 0.000428 0.009723 0.248213
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.920e-04 6.802e-04 0.429 0.66772
## z.lag.1 -6.631e-03 2.147e-03 -3.089 0.00203 **
## tt -2.006e-07 3.875e-07 -0.518 0.60476
## z.diff.lag -3.875e-02 1.814e-02 -2.136 0.03280 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.01873 on 3034 degrees of freedom
## Multiple R-squared: 0.004994, Adjusted R-squared: 0.00401
## F-statistic: 5.076 on 3 and 3034 DF, p-value: 0.00166
##
##
## Value of test-statistic is: -3.0888 3.2828 4.9235
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau3 -3.96 -3.41 -3.12
## phi2 6.09 4.68 4.03
## phi3 8.27 6.25 5.34plot(z4.df3)
NO podemos rechazar la \(H_0\) por lo que parace que existe la raíz unitaria.
Dado el valor crítico de \(\phi_3\) podemos eliminar la tendencia.
z4.df2 <- ur.df(z4, type="drift", selectlags = "BIC")
print(summary(z4.df2))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression drift
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.136136 -0.009342 0.000387 0.009757 0.248276
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.297e-05 3.398e-04 -0.038 0.96955
## z.lag.1 -6.644e-03 2.146e-03 -3.095 0.00198 **
## z.diff.lag -3.865e-02 1.814e-02 -2.131 0.03319 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.01873 on 3035 degrees of freedom
## Multiple R-squared: 0.004906, Adjusted R-squared: 0.00425
## F-statistic: 7.482 on 2 and 3035 DF, p-value: 0.0005738
##
##
## Value of test-statistic is: -3.0954 4.7915
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau2 -3.43 -2.86 -2.57
## phi1 6.43 4.59 3.78plot(z4.df2)
Parece que no hay raíz unitaria por lo que existe la cointegración al igual que regresando inversamente en el primer caso.
La constante parece no ser signiicativa, hay que tener en cuanta que a falta de constante los test se vuelven inestables.
Veamos el modelo sin constante:
z4.df1 <- ur.df(z4, type="none", selectlags = "BIC")
print(summary(z4.df1))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression none
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.136149 -0.009355 0.000374 0.009744 0.248263
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## z.lag.1 -0.006644 0.002146 -3.096 0.00198 **
## z.diff.lag -0.038650 0.018137 -2.131 0.03317 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.01872 on 3036 degrees of freedom
## Multiple R-squared: 0.004906, Adjusted R-squared: 0.004251
## F-statistic: 7.484 on 2 and 3036 DF, p-value: 0.0005724
##
##
## Value of test-statistic is: -3.0959
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau1 -2.58 -1.95 -1.62plot(z4.df1) Parece que no hay raíz unitaria por lo que existe la cointegración
7.1.5 BNP -ASML~5
Primer paso:
e5 <- dynlm(bnp ~ asml)
print(summary(e5))##
## Time series regression with "zoo" data:
## Start = 2005-01-03, End = 2017-04-07
##
## Call:
## dynlm(formula = bnp ~ asml)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.91776 -0.10695 0.03631 0.16570 0.46131
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.449164 0.021901 157.49 <2e-16 ***
## asml 0.067374 0.006143 10.97 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2437 on 3038 degrees of freedom
## Multiple R-squared: 0.03809, Adjusted R-squared: 0.03777
## F-statistic: 120.3 on 1 and 3038 DF, p-value: < 2.2e-16tsdisplay(e5$residuals)
z5 <- e5$residualsLa ACF sugiere que los residuos no son estacionarios. Puede haber una raíz unitaria.
Segundo paso:
Aplicamos la ADF a \(z5\)
- Con constante y tendencia
z5.df3 <- ur.df(z5, type="trend", selectlags = "BIC")
print(summary(z5.df3))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression trend
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.190704 -0.011440 0.000412 0.011774 0.180697
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.641e-04 9.149e-04 0.289 0.77289
## z.lag.1 -5.332e-03 1.881e-03 -2.835 0.00461 **
## tt -6.868e-08 5.217e-07 -0.132 0.89527
## z.diff.lag 1.860e-02 1.815e-02 1.025 0.30566
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02513 on 3034 degrees of freedom
## Multiple R-squared: 0.002897, Adjusted R-squared: 0.001911
## F-statistic: 2.939 on 3 and 3034 DF, p-value: 0.032
##
##
## Value of test-statistic is: -2.8351 2.7242 4.0246
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau3 -3.96 -3.41 -3.12
## phi2 6.09 4.68 4.03
## phi3 8.27 6.25 5.34plot(z5.df3)
NO podemos rechazar la \(H_0\) por lo que parace que existe la raíz unitaria.
Dado el valor crítico de \(\phi_3\) podemos eliminar la tendencia.
z5.df2 <- ur.df(z5, type="drift", selectlags = "BIC")
print(summary(z5.df2))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression drift
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.190661 -0.011482 0.000437 0.011788 0.180748
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.0001596 0.0004558 0.350 0.72621
## z.lag.1 -0.0053115 0.0018739 -2.835 0.00462 **
## z.diff.lag 0.0185846 0.0181505 1.024 0.30596
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02513 on 3035 degrees of freedom
## Multiple R-squared: 0.002892, Adjusted R-squared: 0.002235
## F-statistic: 4.401 on 2 and 3035 DF, p-value: 0.01235
##
##
## Value of test-statistic is: -2.8345 4.079
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau2 -3.43 -2.86 -2.57
## phi1 6.43 4.59 3.78plot(z5.df2)
Parece que hay raíz unitaria por lo que no existe la cointegración
La constante parece no ser signiicativa, hay que tener en cuanta que a falta de constante los test se vuelven inestables.
Veamos el modelo sin constante:
z5.df1 <- ur.df(z5, type="none", selectlags = "BIC")
print(summary(z5.df1))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression none
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.190503 -0.011322 0.000597 0.011948 0.180907
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## z.lag.1 -0.005312 0.001874 -2.835 0.00461 **
## z.diff.lag 0.018626 0.018147 1.026 0.30479
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02512 on 3036 degrees of freedom
## Multiple R-squared: 0.002893, Adjusted R-squared: 0.002236
## F-statistic: 4.405 on 2 and 3036 DF, p-value: 0.0123
##
##
## Value of test-statistic is: -2.8351
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau1 -2.58 -1.95 -1.62plot(z5.df1)
Parece que hay raíz unitaria por lo que existe la cointegración
Aunque hay que subayar que este resultado no es muy fiable por la falta de la estabilidad en la pruba.
7.1.6 BNP -BAYN~6
Primer paso:
e6 <- dynlm(bnp ~ bayn)
print(summary(e6))##
## Time series regression with "zoo" data:
## Start = 2005-01-03, End = 2017-04-07
##
## Call:
## dynlm(formula = bnp ~ bayn)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.90946 -0.10409 0.02965 0.12688 0.45185
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.033638 0.032449 93.49 <2e-16 ***
## bayn 0.164767 0.008145 20.23 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2333 on 3038 degrees of freedom
## Multiple R-squared: 0.1187, Adjusted R-squared: 0.1184
## F-statistic: 409.2 on 1 and 3038 DF, p-value: < 2.2e-16tsdisplay(e6$residuals)
z6 <- e6$residualsLa ACF sugiere que los residuos no son estacionarios. Puede haber una raíz unitaria.
Segundo paso:
Aplicamos la ADF a \(z6\)
- Con constante y tendencia
z6.df3 <- ur.df(z6, type="trend", selectlags = "BIC")
print(summary(z6.df3))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression trend
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.192063 -0.011269 0.000059 0.011536 0.179795
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.096e-04 9.027e-04 0.565 0.57245
## z.lag.1 -5.660e-03 1.950e-03 -2.903 0.00372 **
## tt -2.546e-07 5.176e-07 -0.492 0.62287
## z.diff.lag 2.347e-02 1.815e-02 1.293 0.19612
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02436 on 3034 degrees of freedom
## Multiple R-squared: 0.0032, Adjusted R-squared: 0.002215
## F-statistic: 3.247 on 3 and 3034 DF, p-value: 0.02103
##
##
## Value of test-statistic is: -2.903 2.8449 4.2287
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau3 -3.96 -3.41 -3.12
## phi2 6.09 4.68 4.03
## phi3 8.27 6.25 5.34plot(z6.df3)
NO podemos rechazar la \(H_0\) por lo que parace que existe la raíz unitaria.
Dado el valor crítico de \(\phi_3\) podemos eliminar la tendencia.
z6.df2 <- ur.df(z6, type="drift", selectlags = "BIC")
print(summary(z6.df2))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression drift
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.191870 -0.011348 0.000089 0.011422 0.180088
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.0001225 0.0004420 0.277 0.78173
## z.lag.1 -0.0054412 0.0018981 -2.867 0.00418 **
## z.diff.lag 0.0233333 0.0181493 1.286 0.19867
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02436 on 3035 degrees of freedom
## Multiple R-squared: 0.003121, Adjusted R-squared: 0.002464
## F-statistic: 4.751 on 2 and 3035 DF, p-value: 0.008708
##
##
## Value of test-statistic is: -2.8666 4.1475
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau2 -3.43 -2.86 -2.57
## phi1 6.43 4.59 3.78plot(z6.df2)
Parece que no hay raíz unitaria por lo que existe la cointegración
La constante parece no ser signiicativa, hay que tener en cuanta que a falta de constante los test se vuelven inestables.
Veamos el modelo sin constante:
z6.df1 <- ur.df(z6, type="none", selectlags = "BIC")
print(summary(z6.df1))##
## ###############################################
## # Augmented Dickey-Fuller Test Unit Root Test #
## ###############################################
##
## Test regression none
##
##
## Call:
## lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.191748 -0.011225 0.000212 0.011544 0.180210
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## z.lag.1 -0.005441 0.001898 -2.867 0.00417 **
## z.diff.lag 0.023360 0.018146 1.287 0.19809
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02436 on 3036 degrees of freedom
## Multiple R-squared: 0.003122, Adjusted R-squared: 0.002466
## F-statistic: 4.754 on 2 and 3036 DF, p-value: 0.008678
##
##
## Value of test-statistic is: -2.8672
##
## Critical values for test statistics:
## 1pct 5pct 10pct
## tau1 -2.58 -1.95 -1.62plot(z6.df1)
Parece que no hay raíz unitaria por lo que existe la cointegración